8 networks, distribution and pert/cpm€¦ · · 2016-12-248 networks, distribution and pert/cpm...

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8

Networks, Distribution and PERT/CPM

8.1 What’s Special About Network Models A subclass of models called network LPs warrants special attention for three reasons:

1. They can be completely described by simple, easily understood graphical figures.

2. Under typical conditions, they have naturally integer answers, and one may find a

network LP a useful device for describing and analyzing the various shipment strategies.

3. They are frequently easier to solve than general LPs.

Physical examples that come to mind are pipeline or electrical transmission line networks. Any

enterprise producing a product at several locations and distributing it to many warehouses and/or

customers may find a network LP a useful device for describing and analyzing shipment strategies.

Although not essential, efficient specialized solution procedures may be used to solve network

LPs. These procedures may be as much as 100 times faster than the general simplex method. Bradley,

Brown, and Graves (1977) give a detailed description. Some of these specialized procedures were

developed several years before the simplex method was developed for general LPs.

Figure 8.1 illustrates the network representing the distribution system of a firm using intermediate

warehouses to distribute a product. The firm has two plants (denoted by A and B), three warehouses

(denoted by X, Y, and Z), and four customer areas (denoted by 1, 2, 3, 4). The numbers adjacent to each

node denote the availability of material at that node. Plant A, for example, has nine units available to

be shipped. Customer 3, on the other hand, has 4 units meaning it needs to receive a shipment of four

units.

The number above each arc is the cost per unit shipped along that arc. For example, if five of plant

A’s nine units are shipped to warehouse Y, then a cost of 5 2 = 10 will be incurred as a direct result.

The problem is to determine the amount shipped along each arc, so total costs are minimized and every

customer has his requirements satisfied.

146 Chapter 8 Networks, Distribution & PERT/CPM

Figure 8.1 Three-Level Distribution Network

9

8

1

2

3 1

2

5 7

9 6

7

8 7

4

1

2

3

4

-3

-5

-4

-2

Plants Warehouses Customers

X

Y

Z

B

A

The essential condition on an LP for it to be a network problem is that it be representable as a

network. There can be more than three levels of nodes, any number of arcs between any two nodes,

and upper and lower limits on the amount shipped along a given arc.

With variables defined in an obvious way, the general LP describing this problem is:

[COST] MIN = AX + 2 * AY + 3 * BX + BY + 2 * BZ + 5 * X1

+ 7 * X2 + 9 * Y1 + 6 * Y2 + 7 * Y3 + 8 * Z2 + 7 * Z3

+ 4 * Z4;

[A] AX + AY <= 9;

[B] BX + BY + BZ <= 8;

[X] - AX - BX + X1 + X2 = 0;

[Y] - AY - BY + Y1 + Y2 + Y3 = 0;

[Z] - BZ + Z2 + Z3 + Z4 = 0;

[C1] - X1 - Y1 = -3;

[C2] - X2 - Y2 - Z2 = -5;

[C3] - Y3 - Z3 = -4;

[C4] - Z4 = -2;

There is one constraint for each node that is of a “sources = uses” form. Constraint 5, for example,

is associated with warehouse Y and states that the amount shipped out minus the amount shipped in

must equal 0.

A different view of the structure of a network problem is possible by displaying just the

coefficients of the above constraints arranged by column and row. In the picture below, note that the

apostrophes are placed every third row and column just to help see the regular patterns:

A A B B B X X Y Y Y Z Z Z

X Y X Y Z 1 2 1 2 3 2 3 4

COST: 1 2 3 1 2 5 7 9 6 7 8 7 4 MIN

A: 1 1 ' ' ' ' = 9

B: ' ' 1 1 1 ' ' ' ' ' ' ' ' = 8

X: 1 1 1 1 ' ' =

Y: 1 1 ' 1 1 1 ' =

Z: ' ' ' ' 1 ' ' ' ' ' 1 1 1 =

C1: ' 1 1 ' ' = 3 C2: ' ' 1 1 1 ' = -5

C3: ' ' ' ' ' ' ' ' ' 1 ' 1 ' = -4

C4: ' ' ' ' 1 = 2

Networks, Distribution & PERT/CPM Chapter 8 147

You should notice the key feature of the constraint matrix of a network problem. That is, without

regard to any bound constraints on individual variables, each column has exactly two nonzeroes in the

constraint matrix. One of these nonzeroes is a +1, whereas the other is a 1. According to the

convention we have adopted, the +1 appears in the row of the node from which the arc takes material,

whereas the row of the node to which the arc delivers material is a 1. On a problem of this size, you

should be able to deduce the optimal solution manually simply from examining Figure 8.1. You may

check it with the computer solution below:

Variable Value Reduced Cost

AX 3.000000 0.000000

AY 3.000000 0.000000

BX 0.000000 3.000000

BY 6.000000 0.000000

BZ 2.000000 0.000000

X1 3.000000 0.000000

X2 0.000000 0.000000

Y1 0.000000 5.000000

Y2 5.000000 0.000000

Y3 4.000000 0.000000

Z2 0.000000 3.000000

Z3 0.000000 1.000000

Z4 2.000000 0.000000

Row Slack or Surplus Dual Price

COST 100.000000 -1.000000

A 3.000000 0.000000

B 0.000000 1.000000

X 0.000000 1.000000

Y 0.000000 2.000000

Z 0.000000 3.000000

C1 0.000000 6.000000

C2 0.000000 8.000000

C3 0.000000 9.000000

C4 0.000000 7.000000

This solution exhibits two pleasing features found in the solution to any network problem:

1. If the right-hand side coefficients (the capacities and requirements) are integer, then the

variables will also be integer.

2. If the objective coefficients are integer, then the dual prices will also be integer.

We can summarize network LPs as follows:

1. Associated with each node is a number that specifies the amount of commodity available

at that node (negative implies that commodity is required.)

2. Associated with each arc are:

a) a cost per unit shipped (which may be negative) over the arc,

b) a lower bound on the amount shipped over the arc (typically zero), and

c) an upper bound on the amount shipped over the arc (infinity in our example).

The problem is to determine the flows that minimize total cost subject to satisfying all the supply,

demand, and flow constraints.


8.1.1 Special Cases There are a number of common applications of LP models that are special cases of the standard

network LP. The ones worthy of mention are:

1. Transportation or distribution problems. A two-level network problem, where all the

nodes at the first level are suppliers, all the nodes at the second level are users, and the

only arcs are from suppliers to users, is called a transportation, or distribution model.

2. Shortest and longest path problems. Suppose one is given the road network of the United

States and wishes to find the shortest route from Bangor to San Diego. This is equivalent

to a special case of a network or transshipment problem in which one unit of material is

available at Bangor and one unit is required at San Diego. The cost of shipping over an

arc is the length of the arc. Simple, fast procedures exist for solving this problem. An

important first cousin of this problem, the longest route problem, arises in the analysis of

PERT/CPM projects.

3. The assignment problem. A transportation problem in which the number of suppliers

equals the number of customers, each supplier has one unit available, and each customer

requires one unit, is called an assignment problem. An efficient, specialized procedure

exists for its solution.

4. Maximal flow. Given a directed network with an upper bound on the flow on each arc,

one wants to find the maximum that can be shipped through the network from some

specified origin, or source node, to some other destination, or sink node. Applications

might be to determine the rate at which a building can be evacuated or military material

can be shipped to a distant trouble spot.

8.1.2 Fitting into Network Structure: Roads with No Left Turns The parcel delivery service, UPS got publicity a number of years ago when it claimed that its drivers

did not make left turns. The argument is that at a busy intersection, making a left turn requires the left

turning vehicle to wait for a gap in oncoming traffic. We illustrate here that it sometimes requires a bit

of thought to precisely describe a problem as a network problem. How do we represent restrictions on

turns at an intersection, left turns and U turns in particular? One way of representing turn restrictions

in a standard directed network is to add arcs and nodes to an intersection to represent the valid

possibilities. Figure 8.2 illustrates what one can do to represent a 4-way intersection where left turns

and U turns are prohibited. One node is replaced by 8 nodes and 8 additional arcs. Observe that for a

Roundabout, however, no additional nodes and arcs are required. Some UPS drivers have admitted that

left turns are sometimes made, usually only on streets with low traffic.


Figure 8.2 Network for a 4-Way Intersection with No Left or U Turns

8.2 PERT/CPM Networks and LP Program Evaluation and Review Technique (PERT) and Critical Path Method (CPM) are two closely

related techniques for monitoring the progress of a large project. A key part of PERT/CPM is

calculating the critical path. That is, identifying the subset of the activities that must be performed

exactly as planned in order for the project to finish on time.

We will show that the calculation of the critical path is a very simple network LP problem,

specifically, a longest path problem. You do not need this fact to efficiently calculate the critical path,

but it is an interesting observation that becomes useful if you wish to examine a multitude of

“crashing” options for accelerating a tardy project.


In the table below, we list the activities involved in the simple, but nontrivial, project of building a

house. An activity cannot be started until all of its predecessors are finished:

Activity Predecessors Activity Mnemonic Time (Mnemonic)

Dig Basement DIG 3

Pour Foundation FOUND 4 DIG

Pour Basement Floor POURB 2 FOUND

Install Floor Joists JOISTS 3 FOUND

Install Walls WALLS 5 FOUND

Install Rafters RAFTERS 3 WALLS, POURB

Install Flooring FLOOR 4 JOISTS

Rough Interior ROUGH 6 FLOOR

Install Roof ROOF 7 RAFTERS

Finish Interior FINISH 5 ROUGH, ROOF

Landscape SCAPE 2 POURB, WALLS

In Figure 8.3, we show the so-called PERT (or activity-on-arrow) network for this project. We

would like to calculate the minimum elapsed time to complete this project. Relative to this figure, the

number of interest is simply the longest path from left to right in this figure. The project can be

completed no sooner than the sum of the times of the successive activities on this path. Verify for

yourself that the critical path consists of activities DIG, FOUND, WALLS, RAFTERS, ROOF, and

FINISH and has length 27.

Even though this example can be worked out by hand, almost without pencil and paper, let us

derive an LP formulation for solving this problem. Most people attempting this derivation will come

up with one of two seemingly unrelated formulations.

The first formulation is motivated as follows. Let variables DIG, FOUND, etc. be either 1 or 0

depending upon whether activities DIG, FOUND, etc. are on or not on the critica1 path. The variables

equa1 to one will define the critical path. The objective function will be related to the fact that we want

to find the maximum length path in the PERT diagram.

Our objective is in fact:

MAX = 3 * DIG + 4 * FOUND + 2 * POURB + 3 * JOISTS +

5 * WALLS + 3 * RAFTERS + 4 * FLOOR + 6 *

ROUGH + 7 * ROOF + 5 * FINISH + 2 * SCAPE;


Figure 8.3 Activity-on-Arc PERT/CPM Network

A B C

D

F

H

I

G

E

33

3

4

4

5

5

2

2

6

7

Dig Found

Joists

Scape

Walls

Pour B Rafters

Roof

Finish

RoughFloor

By itself, this objective seems to take the wrong point of view. We do not want to maximize the

project length. However, if we specify the proper constraints, we shall see this objective will seek out

the maximum length path in the PERT network. We want to use the constraints to enforce the

following:

1. DIG must be on the critical path.

2. An activity can be on the critical path only if one of its predecessors is on the critical

path. Further, if an activity is on a critical path, exactly one of its successors must be on

the critical path, if it has successors.

3. Exactly one of SCAPE or FINISH must be on the critical path.

Convince yourself the following set of constraints will enforce the above:

DIG = 1;

FOUND + DIG = 0;

JOISTS — POURB — WALLS + FOUND = 0;

FLOOR + JOISTS = 0;

RAFTERS SCAPE + POURB + WALLS = 0;

ROUGH + FLOOR = 0;

ROOF + RAFTERS = 0;

FINISH + ROUGH + ROOF = 0; + FINISH + SCAPE = +1;

If you interpret the length of each arc in the network as the scenic beauty of the arc, then the

formulation corresponds to finding the most scenic route by which to ship one unit from A to I.


The solution of the problem is:

Optimal solution found at step: 2

Objective value: 27.00000


DIG 1.000000 0.0000000

FOUND 1.000000 0.0000000

POURB 0.0000000 3.000000

JOISTS 0.0000000 0.0000000

WALLS 1.000000 0.0000000

RAFTERS 1.000000 0.0000000

FLOOR 0.0000000 0.0000000

ROUGH 0.0000000 2.000000

ROOF 1.000000 0.0000000

FINISH 1.000000 0.0000000

SCAPE 0.0000000 13.00000


1 27.00000 1.000000

2 0.0000000 6.000000

3 0.0000000 -9.000000

4 0.0000000 -5.000000

5 0.0000000 -2.000000

6 0.0000000 0.0000000

7 0.0000000 2.000000

8 0.0000000 3.000000

9 0.0000000 10.00000

10 0.0000000 15.00000

Notice the variables corresponding to the activities on the critical path have a value of 1. What is

the solution if the first constraint, DIG = 1, is deleted?

It is instructive to look at the PICTURE of this problem in the following figure:

R

J A F

F P O W F F R I S

O O I A T L O R N C

D U U S L E O U O I A

I N R T L R O G O S P

G D B S S S R H F H E

1: 3 4 2 3 5 3 4 6 7 5 2 MAX

2: 1 ' ' ' = 1 3: 1 1 ' ' ' ' ' ' ' ' ' =

4: 1 1 1 1 ' ' =

5: 1 1 ' =

6: ' ' 1 ' 1 1 ' ' ' ' 1 =

7: ' 1 1 ' =

8: ' 1 ' 1 ' =

9: ' ' ' ' ' ' ' 1 1 1 ' =

10: ' ' 1 1 = 1

Notice that each variable has at most two coefficients in the constraints. When two, they are +1

and 1. This is the distinguishing feature of a network LP.


Now, let us look at the second possible formulation. The motivation for this formulation is to

minimize the elapsed time of the project. To do this, realize that each node in the PERT network

represents an event (e.g., as follows: A, start digging the basement; C, complete the foundation; and I,

complete landscaping and finish interior).

Define variables A, B, C, …, H, I as the time at which these events occur. Our objective function is

then:

MIN = I A;

These event times are constrained by the fact that each event has to occur later than each of its

preceding events, at least by the amount of any intervening activity. Thus, we get one constraint for

each activity:

B A >= 3; ! DIG;

C B >= 4; ! FOUND;

E C >= 2;

D C >= 3;

E C >= 5;

F D >= 4;

G E >= 3;

H F >= 6;

H G >= 7;

I H >= 5;

I E >= 2;

The solution to this problem is:




I 27.00000 0.0000000

A 0.0000000 0.0000000

B 3.000000 0.0000000

C 7.000000 0.0000000

E 12.00000 0.0000000

D 10.00000 0.0000000

F 14.00000 0.0000000

G 15.00000 0.0000000

H 22.00000 0.0000000


1 27.00000 1.000000

2 0.0000000 -1.000000

3 0.0000000 -1.000000

4 3.000000 0.0000000

5 0.0000000 0.0000000

6 0.0000000 -1.000000

7 0.0000000 0.0000000

8 0.0000000 -1.000000

9 2.000000 0.0000000

10 0.0000000 -1.000000

11 0.0000000 -1.000000

12 13.00000 0.0000000


Notice that the objective function value equals the critical path length. We can indirectly identify

the activities on the critical path by noting the constraints with nonzero dual prices. The activities

corresponding to these constraints are on the critical path. This correspondence makes sense. The

right-hand side of a constraint is the activity time. If we increase the time of an activity on the critical

path, it should increase the project length and thus should have a nonzero dual price. What is the

solution if the first variable, A, is deleted?

The PICTURE of the coefficient matrix for this problem follows:

A B C D E F G H I

1:—1 ' ' 1 MIN

2:1 1 ' ' 3

3: ' 1 '1 ' ' ' ' 4

4: 1 ' 1 ' 2

5: 1 1 ' > 3

6: ' 1 ' 1 ' ' ' 5

7: 1 1 ' > 4

8: ' 1 1 3

9: ' ' ' 1 ' 1 ' 6

10: ' 1 1 > 7

11: ' ' 1 1 5

12: ' ' ' 1 ' ' '1 2

Notice the PICTURE of this formulation is essentially the PICTURE of the previous formulation

rotated ninety degrees. Even though these two formulations originally were seemingly unrelated, there

is really an incestuous relationship between the two, a relationship that mathematicians politely refer to

as duality.

8.3 Activity-on-Arc vs. Activity-on-Node Network Diagrams Two conventions are used in practice for displaying project networks: (1) Activity-on-Arc (AOA) and

(2) Activity-on-Node (AON). Our previous example used the AOA convention. The characteristics of

the two are:

AON

Each activity is represented by a node in the network.

A precedence relationship between two activities is represented by an arc or link between

the two.

AON may be less error prone because it does not need dummy activities or arcs.

AOA

Each activity is represented by an arc in the network.

If activity X must precede activity Y, there are X leads into arc Y. The nodes thus

represent events or “milestones” (e.g., “finished activity X”). Dummy activities of zero

length may be required to properly represent precedence relationships.

AOA historically has been more popular, perhaps because of its similarity to Gantt charts

used in scheduling.

An AON project with six activities is shown in Figure 8.4. The number next to each node is the

duration of the activity. Activities A and B are the sources or start of the project. Activity F is the final

activity. By inspection, you can discover that the longest path consists of activities A, C, E, and F. It


has a length of 29. The corresponding AOA network for the same project is shown in Figure 8.5. In the

AOA network, we have enclosed the activity letters in circles above the associated arc. The unenclosed

numbers below each arc are the durations of the activities. We have given the nodes, or milestones,

arbitrary number designations enclosed in squares. Notice the dummy activity (the dotted arc) between

nodes 3 and 4. This is because a dummy activity will be required in an AOA diagram anytime that two

activities (e.g., A and B) share some (e.g., activity D), but not all (e.g., activity C), successor activities.

Figure 8.4 An Activity-on-Node Representation

7

9 5 6

9 8 F D B

A C E

Figure 8.5 An Activity-on-Arc Representation

0

A

D

C

E

F

9 5

6

9 8

7 B

7

3 1 5

6 4 2

8.4 Crashing of Project Networks Once the critical path length for a project has been identified, the next question invariably asked is: can

we shorten the project? The process of decreasing the duration of a project or activity is commonly

called crashing. For many construction projects, it is common for the customer to pay an incentive to

the contractor for finishing the project in a shorter length of time. For example, in highway repair

projects, it is not unusual to have incentives from $5,000 to $25,000 per day that the project is finished

before a target date.


8.4.1 The Cost and Value of Crashing There is value in crashing a project. In order to crash a project, we must crash one or more activities.

Crashing an activity costs money. Deciding to crash an activity requires us to compare the cost of

crashing that activity with the value of the resulting reduction in project length. This decision is

frequently complicated by the fact that some negotiation may be required between the party that incurs

the cost of crashing the activity (e.g., the contractor) and the party that enjoys the value of the crashed

project (e.g., the customer).

8.4.2 The Cost of Crashing an Activity An activity is typically crashed by applying more labor to it (e g., overtime or a second shift). We

might typically expect that using second-shift labor could cost 1.5 times as much per hour as first-shift

labor. We might expect third-shift labor to cost twice as much as first-shift labor.

Consider an activity that can be done in six days if only first-shift labor is used and has a labor

cost of $6,000. If we allow the use of second-shift labor and thus work two shifts per day, the activity

can be done in three days for a cost of 3 1000 + 3 l000 1.5 = 7,500. If third-shift labor is allowed,

then the project can be done in two days by working three shifts per day and incurring a total of:

2 1000 + 2 1000 1.5 + 2 1000 2 = $9,000.

Thus, we get a crashing cost curve for the activity as shown in Figure 8.6:

Figure 8.6 Activity Crash Cost Curve

Activity Duration

Cost

1.5

1.25

1

1/3 1/2 1

Normal time

8.4.3 The Value of Crashing a Project There are two approaches to deciding upon the amount of project crashing: (a) we simply specify a

project duration time and crash enough to achieve this duration, or (b) we estimate the value of

crashing it for various days. As an example of (a), in 1987 a new stadium was being built for the

Montreal Expos baseball team. The obvious completion target was the first home game of the season.


As an example of (b), consider an urban expressway repair. What is the value per day of

completing it early? Suppose that 6,000 motorists are affected by the repair project and each is delayed

by 10 minutes each day because of the repair work (e.g., by taking alternate routes or by slower

traffic). The total daily delay is 6,000 10 = 60,000 minutes = 1000 hours. If we assign an hourly cost

of $5/person hours, the social value of reducing the repair project by one day is $5,000.

8.4.4 Formulation of the Crashing Problem Suppose we have investigated the crashing possibilities for each activity or task in our previous project

example. These estimates are summarized in the following table:

Minimum duration Normal duration if crashed

Activity Predecessor (Days) (Days) $/Day

A — 9 5 5000

B — 7 3 6000

C A 5 3 4000

D A,B 8 4 2000

E C 6 3 3000

F D,E 9 5 9000

For example, activity A could be done in five days rather than nine. However, this would cost us

an extra (9 5) 5000 = $20,000.

First, consider the simple case where we have a hard due date by which the project must be done.

Let us say 22 days in this case. How would we decide which activities to crash? Activity D is the

cheapest to crash per day. However, it is not on the critical path, so its low cost is at best just

interesting.

Let us define:

EFi = earliest finish time of activity i, taking into account any crashing that is done;

Ci = number of days by which activity i is crashed.

In words, the LP model will be:

Minimize Cost of crashing

subject to

For each activity j and each predecessor i:

earliest finish of j earliest finish of predecessor i + actual duration of j;

For each activity j:

minimum duration for j if crashed actual duration of j normal duration for j.

A LINGO formulation is:

! Find optimal crashing for a project with a due date;

SETS:

TASK: NORMAL, FAST, COST, EF, ACTUAL;

PRED( TASK, TASK):;

ENDSETS


DATA:

TASK, NORMAL, FAST, COST =

A 9 5 5000

B 7 3 6000

C 5 3 4000

D 8 4 2000

E 6 3 3000

F 9 5 9000;

PRED =

A, C

A, D

B, D

C, E

D, F

E, F;

DUEDATE = 22;

ENDDATA

!-------------------------------------;

! Minimize the cost of crashing;

[OBJ] MIN = @SUM( TASK( I): COST( I)*( NORMAL( I) - ACTUAL( I)));

! For tasks with no predecessors...;

@FOR( TASK( J): EF( J) >= ACTUAL( J););

! and for those with predecessors;

@FOR( PRED( I, J):

EF( J) >= EF( I) + ACTUAL( J);

);

! Bound the actual time;

@FOR( TASK( I):

@BND( FAST(I), ACTUAL( I), NORMAL( I));

);

! Last task is assumed to be last in project;

EF( @SIZE( TASK)) <= DUEDATE;

Part of the solution is:

Global optimal solution found at step: 24



EF( A) 7.000000 0.0000000

EF( B) 7.000000 0.0000000

EF( C) 10.00000 0.0000000

EF( D) 13.00000 0.0000000

EF( E) 13.00000 0.0000000

EF( F) 22.00000 0.0000000

ACTUAL( A) 7.000000 0.0000000

ACTUAL( B) 7.000000 -4000.000

ACTUAL( C) 3.000000 1000.000

ACTUAL( D) 6.000000 0.0000000

ACTUAL( E) 3.000000 2000.000

ACTUAL( F) 9.000000 -2000.000

Thus, for an additional cost of $31,000, we can meet the 22-day deadline.


Now, suppose there is no hard project due date, but we do receive an incentive payment of $5000

for each day we reduce the project length. Define PCRASH = number of days the project is finished

before the twenty-ninth day. Now, the formulation is:

! Find optimal crashing for a project with

a due date and incentive for early completion;

SETS:

TASK: NORMAL, FAST, COST, EF, ACTUAL;

PRED( TASK, TASK):;

ENDSETS

DATA:

TASK, NORMAL, FAST, COST =

A 9 5 5000

B 7 3 6000

C 5 3 4000

D 8 4 2000

E 6 3 3000

F 9 5 9000;

PRED =

A, C

A, D

B, D

C, E

D, F

E, F;

! Incentive for each day we beat the due date;

INCENT = 5000;

DUEDATE = 29;

ENDDATA

!-------------------------------------;

! Minimize the cost of crashing

less early completion incentive payment;

[OBJ] MIN = @SUM( TASK( I): COST( I)*( NORMAL( I) - ACTUAL( I)))

- INCENT * PCRASH;

! For tasks with no predecessors...;

@FOR( TASK( J): EF( J) >= ACTUAL( J););

! and for those with predecessors;

@FOR( PRED( I, J):

EF( J) >= EF( I) + ACTUAL( J);

);

! Bound the actual time;

@FOR( TASK( I):

@BND( FAST(I), ACTUAL( I), NORMAL( I));

);

! Last task is assumed to be last in project;

EF( @SIZE( TASK)) + PCRASH = DUEDATE;


From the solution, we see we should crash it by five days to give a total project length of

twenty-four days:

Global optimal solution found at step: 21

Objective value: -6000.000


PCRASH 5.000000 0.0000000

EF( A) 7.000000 0.0000000

EF( B) 7.000000 0.0000000

EF( C) 12.00000 0.0000000

EF( D) 15.00000 0.0000000

EF( E) 15.00000 0.0000000

EF( F) 24.00000 0.0000000

ACTUAL( A) 7.000000 0.0000000

ACTUAL( B) 7.000000 -6000.000

ACTUAL( C) 5.000000 -1000.000

ACTUAL( D) 8.000000 0.0000000

ACTUAL( E) 3.000000 0.0000000

ACTUAL( F) 9.000000 -4000.000

The excess of the incentive payments over crash costs is $6,000.

8.5 Resource Constraints in Project Scheduling For many projects, a major complication is that there are a limited number of resources. The limited

resources require you to do tasks individually that otherwise might be done simultaneously. Pritzker,

Watters, and Wolfe (1969) gave a formulation representing resource constraints in project and jobshop

scheduling problems. The formulation is based on the following key ideas: a) time is discrete rather

than continuous (e.g., each period is a day), b) for every activity and every discrete period there is a 0/1

variable that is one if that activity starts in that period, and c) for every resource and period there is a

constraint that enforces the requirement that the amount of resource required in that period does not

exceed the amount available.

To illustrate, we take the example considered previously with shorter activity times, so the total

number of periods is smaller:

MODEL:

! PERT/CPM project scheduling with resource constraints(PERTRSRC);

! There is a limited number of each resource/machine.

! An activity cannot be started until: 1) all its predecessors have

completed, and 2) resources/machines required are available.;

SETS:

! There is a set of tasks with a given duration, and

a start time to be determined;

TASK: TIME, START, ES;

! The precedence relations, the first task in the

precedence relationship needs to be completed before the

second task can be started;

PRED( TASK, TASK);

! There are a set of periods;

PERIOD;


RESOURCE: CAP;

! Some operations need capacity in some department;

TXR( TASK, RESOURCE): NEED;

! SX( I, T) = 1 if task I starts in period T;

TXP( TASK, PERIOD): SX;

RXP( RESOURCE, PERIOD);

ENDSETS

DATA:

! Upper limit on number of periods required to complete the project;

PERIOD = 1..20;

! Task names and duration;

TASK TIME =

FIRST 0

FCAST 7

SURVEY 2

PRICE 1

SCHED 3

COSTOUT 2

FINAL 4;

! The predecessor/successor combinations;

PRED= FIRST,FCAST, FIRST,SURVEY,

FCAST,PRICE, FCAST,SCHED, SURVEY,PRICE,

SCHED,COSTOUT, PRICE,FINAL, COSTOUT,FINAL;

! There are 2 departments, accounting and operations,

with capacities...;

RESOURCE = ACDEPT, OPNDEPT;

CAP = 1, 1;

! How much each task needs of each resource;

TXR, NEED =

FCAST, OPNDEPT, 1

SURVEY, OPNDEPT, 1

SCHED, OPNDEPT, 1

PRICE, ACDEPT, 1

COSTOUT, ACDEPT, 1;

ENDDATA

!----------------------------------------------------------;

! Minimize start time of last task;

MIN = START( @SIZE( TASK));

! Start time for each task;

@FOR( TASK( I):

[DEFSTRT] START( I) = @SUM( PERIOD( T): T * SX( I, T));

);

@FOR( TASK( I):

! Each task must be started in some period;

[MUSTDO] @SUM( PERIOD( T): SX( I, T)) = 1;

! The SX vars are binary, i.e., 0 or 1;

@FOR( PERIOD( T): @BIN( SX( I, T)););

);

! Precedence constraints;

@FOR( PRED( I, J):

[PRECD] START( J) >= START( I) + TIME( I);


);

! Resource usage, For each resource R and period T;

@FOR( RXP( R, T):

! Sum over all tasks I that use resource R in period T;

[RSRUSE] @SUM( TXR( I, R):

@SUM( PERIOD( S)| S #GE# ( T - ( TIME( I) - 1)) #AND# S #LE# T:

NEED( I, R) * SX( I, S))) <= CAP( R);

);

! The following makes the formulation tighter;

! Compute earliest start disregarding resource constraints;

@FOR( TASK( J):

ES( J) = @SMAX( 0, @MAX( PRED( I, J): ES( I) + TIME(I)));

! Task cannot start earlier than unconstrained early start;

@SUM( PERIOD(T) | T #LE# ES( J): SX( J, T)) = 0;

);

END

When solved, we get a project length of 14. If there were no resource constraints, then the project

length would be 13:

Global optimal solution found


Variable Value

START( FIRST) 1.000000

START( FCAST) 1.000000

START( SURVEY) 11.00000

START( PRICE) 13.00000

START( SCHED) 8.000000

START( COSTOUT) 11.00000

START( FINAL) 14.00000

8.6 Path Formulations In many network problems, it is natural to think of a solution in terms of paths that material takes as it

flows through the network. For example, in Figure 8.1, there are thirteen possible paths. Namely:

A X 1, A X 2, A Y 1, A Y 2, A Y 3, B X 1, B X 2,

B Y 1, B Y 2, B Y 3, B Z 2, B Z 3,

B Z 4

One can, in fact, formulate decision variables in terms of complete paths rather than just simple

links, where the path decision variable corresponds to using a combination of links. This is a form of

what is sometimes called a composite variable approach. The motivations for using the path approach

are:

1. More complicated cost structures can be represented. For example, Geoffrion and Graves

(1974) use the path formulation to represent “milling in transit” discount fare structures

in shipping food and feed products.


2. Path-related restrictions can be incorporated. For example, regulations allow a truck

driver to be on duty for at most 10 hours. Thus, in a truck routing network one would not

consider paths longer than 10 hours. In a supply chain network, a path that is long may be

prohibited because it may cause lead times to be too long.

3. The number of rows (constraints) in the model may be substantially less.

4. In integer programs where some, but not all, of the problem has a network structure, the

path formulation may be easier to solve.

8.6.1 Example Let us reconsider the first problem (Figure 8.1, page 146). Suppose shipments from A to X are made by

the same carrier as shipments from X to 2. This carrier will give a $1 per unit “milling-in-transit”

discount for each unit it handles from both A to X and X to 2. Further, the product is somewhat fragile

and cannot tolerate a lot of transportation. In particular, it cannot be shipped both over link BX and

X2 or both over links AY and Y1.

Using the notation AX1 = number of units shipped from A to X to 1, etc., the path formulation is:

MIN = 6 * PAX1 + 7 * PAX2 + 8 * PAY2

+ 9 * PAY3 + 8 * PBX1 + 10 * PBY1

+ 7 * PBY2 + 8 * PBY3 + 10 * PBZ2

+ 9 * PBZ3 + 6 * PBZ4;

[A] PAX1 + PAX2 + PAY2 + PAY3 <= 9;

[B] PBX1 + PBY1 + PBY2 + PBY3

+ PBZ2 + PBZ3 + PBZ4 <= 8;

[C1] PAX1 + PBX1 + PBY1 = 3;

[C2] PAX2 + PAY2 + PBY2 + PBZ2 = 5;

[C3] PAY3 + PBY3 + PBZ3 = 4;

[C4] PBZ4 = 2;

Notice the cost of path AX2 = 1 + 7 1 = 7. In addition, paths BX2 and AY1 do not appear. This

model has only six constraints as opposed to nine in the original formulation. The reduction in

constraints arises from the fact that, in path formulations, one does not need the “sources = uses”

constraints for intermediate nodes.

In general, the path formulation will have fewer rows, but more decision variables than the

corresponding network LP model.

When we solve, we get:

Objective value= 97.0000

Variable Value

PAX1 3.000000

PAX2 3.000000

PBY2 2.000000

PBY3 4.000000

PBZ4 2.000000

This is cheaper than the previous solution, because the three units shipped over path AX2 go for $1

per unit less.

A path formulation need not have a naturally integer solution. If the path formulation, however, is

equivalent to a network LP, then it will have a naturally integer solution.

The path formulation is popular in long-range forest planning. See, for example, Davis and

Johnson (1986), where it is known as the “Model I” approach. The standard network LP based


formulation is known as the “Model II” approach. In a forest planning Model II, a link in the network

represents a decision to plant an acre of a particular kind of tree in some specified year and harvest it in

some future specified year. A node represents a specific harvest and replant decision. A decision

variable in Model I is a complete prescription of how to manage (i.e., harvest and replant) a given

piece of land over time. Some Model I formulations in forest planning may have just a few hundred

constraints, but over a million decision variables or paths.

There is a generalization of the path formulation to arbitrary linear programs, known as

Fourier/Motzkin/Dines elimination, see for example Martin (1999) and Dantzig (1963). The

transformation of a network LP to the path formulation involves eliminating a particular node

(constraint), by generating a new variable for every combination of input arc and output arc incident to

the node. A constraint in an arbitrary LP can be eliminated if it is first converted to a constraint with a

right-hand side of zero and then a new variable is generated for every combination of positive and

negative coefficient in the constraint. The disadvantage of this approach is that even though the

number of constraints is reduced to one, the number of variables may grow exponentially with the

number of original constraints.

A variable corresponding to a path in a network is an example of a composite variable, a general

approach that is sometimes useful for representing complicated/ing constraints. A composite variable

is one that represents a feasible combination of two or more original variables. The complicating

constraints are represented implicitly by generating only those composite variables that correspond to

feasible combinations and values of the original variables.

8.7 Path Formulations of Undirected Networks In many communications networks, the arcs have capacity, but are undirected. For example, when you

are carrying on a phone conversation with someone in a distant city, the conversation uses capacity on

all the links in your connection. However, you cannot speak of a direction of flow of the connection.

A major concern for a long distance communications company is the management of its

communications network. This becomes particularly important during certain holidays, such as

Mother’s Day. Not only does the volume of calls increase on these days, but also the pattern of calls

changes dramatically from the business-oriented traffic during weekdays in the rest of the year. A

communications company faces two problems: (a) the design problem. That is, what capacity should

be installed on each link? As well as, (b) the operations problem. That is, given the installed capacity,

how are demands best routed? The path formulation is an obvious format for modeling an undirected

network. The following illustrates the operational problem.


Consider the case of a phone company with the network structure shown in Figure 8.7:

Figure 8.7 Phone Company Network Structure

SEACHI

DNV ATL

MIA

80

95

110

200

105

The number next to each arc is the number of calls that can be in progress simultaneously along

that arc. If someone in MIA tries to call his mother in SEA, the phone company must first find a path

from MIA to SEA such that each arc on that path is not at capacity. It is quite easy to inefficiently use

the capacity. Suppose there is a demand for 110 calls between CHI and DNV and 90 calls between ATL

and SEA. Further, suppose all of these calls were routed over the ATL, DNV link. Now, suppose we

wish to make a call between MIA and SEA. Such a connection is impossible because every path

between the two contains a saturated link (i.e., either ATL, DNV or CHI, ATL). However, if some of the

110 calls between CHI and DNV were routed over the CHI, SEA, DNV links, then one could make calls

between MIA and SEA. In conventional voice networks, a call cannot be rerouted once it has started. In

packet switched data networks and, to some extent, in cellular phone networks, some rerouting is

possible.

8.7.1 Example Suppose during a certain time period the demands in the table below occur for connections between

pairs of cities:

DNV CHI ATL MIA

SEA 10 20 38 33

DNV 42 48 23

CHI 90 36

ATL 26

Which demands should be satisfied and via what routes to maximize the number of connections

satisfied?


Solution. If we use the path formulation, there will be two paths between every pair of cities

except ATL and MIA. We will use the notation P1ij for number of calls using the shorter or more

northerly path between cities i and j, and P2ij for the other path, if any. There will be two kinds of

constraints:

1) a capacity constraint for each link, and

2) an upper limit on the calls between each pair of cities, based on available demand.

A formulation is:

! Maximize calls carried;

MAX = P1MIAATL + P1MIADNV + P2MIADNV

+ P1MIASEA + P2MIASEA + P1MIACHI

+ P2MIACHI + P1ATLDNV + P2ATLDNV

+ P1ATLSEA + P2ATLSEA + P1ATLCHI

+ P2ATLCHI + P1DNVSEA + P2DNVSEA

+ P1DNVCHI + P2DNVCHI + P1SEACHI

+ P2SEACHI;

! Capacity constraint for each link;

[KATLMIA] P1MIAATL + P1MIADNV + P2MIADNV

+ P1MIASEA + P2MIASEA + P1MIACHI

+ P2MIACHI <= 105;

[KATLDNV] P1MIADNV + P1MIASEA + P1MIACHI

+ P1ATLDNV + P1ATLSEA + P1ATLCHI

+ P2DNVSEA + P2DNVCHI + P2SEACHI <= 200;

[KDNVSEA] P2MIADNV + P1MIASEA + P1MIACHI



[KSEACHI] P2MIADNV + P2MIASEA + P1MIACHI



[KATLCHI] P2MIADNV + P2MIASEA + P2MIACHI



! Demand constraints for each city pair;

[DMIAATL] P1MIAATL <= 26;

[DMIADNV] P1MIADNV + P2MIADNV <= 23;

[DMIASEA] P1MIASEA + P2MIASEA <= 33;

[DMIACHI] P1MIACHI + P2MIACHI <= 36;

[DATLDNV] P1ATLDNV + P2ATLDNV <= 48;

[DATLSEA] P1ATLSEA + P2ATLSEA <= 38;

[DATLCHI] P1ATLCHI + P2ATLCHI <= 90;

[DDNVSEA] P1DNVSEA + P2DNVSEA <= 10;

[DDNVCHI] P1DNVCHI + P2DNVCHI <= 42;

[DSEACHI] P1SEACHI + P2SEACHI <= 20;


When this formulation is solved, we see we can handle 322 out of the total demand of 366 calls:




P1MIAATL 26.00000 0.000000

P1MIADNV 23.00000 0.000000

P2MIADNV 0.00000 2.000000

P1MIASEA 0.00000 0.000000

P2MIASEA 0.00000 0.000000

P1MIACHI 25.00000 0.000000

P2MIACHI 0.00000 0.000000

P1ATLDNV 48.00000 0.000000

P2ATLDNV 0.00000 2.000000

P1ATLSEA 38.00000 0.000000

P2ATLSEA 0.00000 0.000000

P1ATLCHI 23.00000 0.000000

P2ATLCHI 67.00000 0.000000

P1DNVSEA 3.50000 0.000000

P2DNVSEA 6.50000 0.000000

P1DNVCHI 5.50000 0.000000

P2DNVCHI 36.50000 0.000000

P1SEACHI 20.00000 0.000000

P2SEACHI 0.00000 2.000000


1 322.00000 1.000000

KATLMIA 31.00000 0.000000

KATLDNV 0.00000 0.000000

KDNVSEA 0.00000 1.000000

KSEACHI 0.00000 0.000000

KATLCHI 0.00000 1.000000

DMIAATL 0.00000 1.000000

DMIADNV 0.00000 1.000000

DMIASEA 33.00000 0.000000

DMIACHI 11.00000 0.000000

DATLDNV 0.00000 1.000000

DATLSEA 0.00000 0.000000

DATLCHI 0.00000 0.000000

DDNVSEA 0.00000 0.000000

DDNVCHI 0.00000 0.000000

DSEACHI 0.00000 1.000000

Verify that the demand not carried is MIA-CHI: 11 and MIA-SEA: 33. Apparently, there are a

number of alternate optima.

8.8 Double Entry Bookkeeping: A Network Model of the Firm Authors frequently like to identify who was the first to use a given methodology. A contender for the

distinction of formulating the first network model is Fra Luca Pacioli. In 1594, while director of a

Franciscan monastery in Italy, he published a description of the accounting convention that has come

to be known as double entry bookkeeping. From the perspective of networks, each double entry is an

arc in a network.


To illustrate, suppose you start up a small dry goods business. During the first two weeks, the

following transactions occur:

CAP 1) You invest $50,000 of capital in cash to start the business.

UR 2) You purchase $27,000 of product on credit from supplier S.

PAY 3) You pay $13,000 of your accounts payable to supplier S.

SEL 4) You sell $5,000 of product to customer C for $8,000 on credit.

REC 5) Customer C pays you $2,500 of his debt to you.

In our convention, liabilities and equities will typically have negative balances. For example, the

initial infusion of cash corresponds to a transfer (an arc) from the equity account (node) to the cash

account, with a flow of $50,000. The purchase of product on credit corresponds to an arc from the

accounts payable account node to the raw materials inventory account, with a flow of $27,000. Paying

$13,000 to the supplier corresponds to an arc from the cash account to the accounts payable account,

with a flow of $13,000. Figure 8.8 illustrates.

Figure 8.8 Double Entry Bookkeeping as a Network Model

Cash

Retained

earnings

Accounts

receivable

Accounts

payable

Rawmaterial

inventory

Equity50,000

13,000

27,000

5,000

2,5

00

8,000

8.9 Extensions of Network LP Models There are several generalizations of network models that are important in practice. These extensions

share two features in common with true network LP models, namely:

They can be represented graphically.

Specialized, fast solution procedures exist for several of these generalizations.

The one feature typically not found with these generalizations is:

Solutions are usually not naturally integer, even if the input data are integers.


The important generalizations we will consider are:

1. Networks with Gains. Sometimes called generalized networks, this generalization allows

a specified gain or loss of material as it is shipped from one node to another. Structurally,

these problems are such that every column has at most two nonzeroes in the constraint

matrix. However, the requirement that these coefficients be +1 and 1 is relaxed.

Specialized procedures, which may be twenty times faster than the regular simplex

method, exist for solving these problems.

Examples of “shipments” with such gains or losses are: investment in an

interest-bearing account, electrical transmission with loss, natural gas pipeline shipments

where the pipeline pumps burn natural gas from the pipeline, and work force attrition.

Stroup and Wollmer (1992) show how a network with gains model is useful in the airline

industry for deciding where to purchase fuel and where to ferry fuel from one stop to

another. Truemper (1976) points out, if the network with gains has no circuits when

considered as an undirected network, then it can be converted to a pure network model by

appropriate scaling.

2. Undirected Networks. In communications networks, there is typically no direction of

shipment. The arcs are undirected.

3. Multicommodity Networks. In many distribution situations, there are multiple

commodities moving through the network, all competing for scarce network capacity.

Each source may produce only one of the commodities and each destination, or sink, may

accept only one specific commodity.

4. Leontief Flow. In a so-called Leontief input-output model (see Leontief, 1951), each

activity uses several commodities although it produces only one commodity. For

example, one unit of automotive production may use a half ton of steel, 300 pounds of

plastic, and 100 pounds of glass. Material Requirements Planning (MRP) models have

the same feature. If each output required only one input, then we would simply have a

network with gains. Special purpose algorithms exist for solving Leontief Flow and MRP

models. See, for example, Jeroslow, Martin, Rardin, and Wang (1992).

5. Activity/Resource Diagrams. If Leontief flow models are extended, so each activity can

have not only several inputs, but also several outputs, then one can in fact represent

arbitrary LPs. We call the obvious extension of the network diagrams to this case an

activity/resource diagram.

8.9.1 Multicommodity Network Flows In a network LP, one assumption is a customer is indifferent, except perhaps for cost, to the source

from which his product was obtained. Another assumption is that there is a single commodity flowing

through the network. In many network-like situations, there are multiple distinct commodities flowing

through the network. If each link has infinite capacity, then an independent network flow LP could be

solved for each commodity. However, if a link has a finite capacity that applies to the sum of all

commodities flowing over that link, then we have a multicommodity network problem.

The most common setting for multicommodity network problems is in shipping. The network

might be a natural gas pipeline network and the commodities might be different fuels shipped over the

network. In other shipping problems, such as traffic assignment or overnight package delivery, each

origin/destination pair constitutes a commodity.

The crucial feature is identity of the commodities must be maintained throughout the network. That

is, customers care which commodity gets delivered. An example is a metals supply company that ships


aluminum bars, stainless steel rings, steel beams, etc., all around the country, using a single limited

capacity fleet of trucks.

In general form, the multicommodity network problem is defined as:

Dik = demand for commodity k at node i, with negative values denoting supply;

Cijk = cost per unit of shipping commodity k from node i to node j;

Uij = capacity of the link from node i to node j.

We want to find:

Xijk = amount of commodity k shipped from node i to node j, so as to:

min kji

cijk xijk

subject to:

For each commodity k and node t :

i

xitk = Dtk + j

xtjk

For each link i, j:

k

xijk Uij

8.9.2 Reducing the Size of Multicommodity Problems If the multiple commodities correspond to origin destination pairs and the cost of shipping a unit over a

link is independent of the final destination, then you can aggregate commodities over destinations.

That is, you need identify a commodity only by its origin, not by both origin and destination. Thus, you

have as many commodities as there are origins, rather than (number of origins) (number of

destinations). For example, in a 100-city problem, using this observation, you would have only 100

commodities, rather than 10,000 commodities.

One of the biggest examples of multicommodity network problems in existence are the Patient

Distribution System models developed by the United States Air Force for planning for transport of sick

or wounded personnel.

8.9.3 Multicommodity Flow Example You have decided to compete with Federal Express by offering “point to point” shipment of materials.

Starting small, you have identified six cities as the ones you will first serve. The matrix below

represents the average number of tons potential customers need to move between each

origin/destination pair per day. For example, people in city 2 need to move four tons per day to city 3:

Demand in tons, D(i, j),

by O/D pair

Cost/ton shipped, C(i, j), by link

Capacity in tons, U(i, j), By link

To: 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6

From

1 0 5 9 7 0 4 0 4 5 8 9 9 0 2 3 2 1 20

2 0 0 4 0 1 0 3 0 3 2 4 6 0 0 2 8 3 9

3 0 0 0 0 0 0 5 3 0 2 3 5 3 0 0 1 3 9

4 0 0 0 0 0 0 7 3 3 0 5 6 5 4 6 0 5 9

5 0 4 0 2 0 8 8 5 3 6 0 3 1 0 2 7 0 9

6 0 0 0 0 0 0 9 7 4 5 5 0 9 9 9 9 9 0


Rather than use a hub system as Federal Express does, you will ship the materials over a regular

directed network. The cost per ton of shipping from any node i to any other node j is denoted by C(i, j).

There is an upper limit on the number of tons shipped per day over any link in the network of U(i, j).

This capacity restriction applies to the total amount of all goods shipped over that link, regardless of

origin or destination. Note U(i, j) and C(i, j) apply to links in the network, whereas D(i, j) applies to

origin/destination pairs. This capacity restriction applies only to the directed flow. That is, U(i, j) need

not equal U(j, i). It may be that none of the goods shipped from origin i to destination j moves over

link (i, j). It is important goods maintain their identity as they move through the network. Notice city 6

looks like a hub. It has high capacity to and from all other cities.

In order to get a compact formulation, we note only three cities, 1, 2, and 5, are suppliers. Thus,

we need keep track of only three commodities in the network, corresponding to the city of origin for

the commodity. Define:

Xijk = tons shipped from city i to city j of commodity k.

The resulting formulation is:

MODEL:

! Keywords: multi-commodity, network flow, routing;

! Multi-commodity network flow problem;

SETS:

! The nodes in the network;

NODES/1..6/:;

! The set of nodes that are origins;

COMMO(NODES)/1, 2, 5/:;

EDGES(NODES, NODES): D, C, U, V;

NET(EDGES, COMMO): X;

ENDSETS

DATA:

! Demand: amount to be shipped from

origin(row) to destination(col);

D = 0 5 9 7 0 4

0 0 4 0 1 0

0 0 0 0 0 0

0 0 0 0 0 0

0 4 0 2 0 8

0 0 0 0 0 0;

! Cost per unit shipped over a arc/link;

C = 0 4 5 8 9 9

3 0 3 2 4 6

5 3 0 2 3 5

7 3 3 0 5 6

8 5 3 6 0 3

9 7 4 5 5 0;

! Upper limit on amount shipped on each link;

U = 0 2 3 2 1 20

0 0 2 8 3 9

3 0 0 1 3 9

5 4 6 0 5 9

1 0 2 7 0 9

9 9 9 9 9 0;


! Whether an arc/link exists or not;

! V = 0 if U = 0;

! V = 1 otherwise;

V = 0 1 1 1 1 1

0 0 1 1 1 1

1 0 0 1 1 1

1 1 1 0 1 1

1 0 1 1 0 1

1 1 1 1 1 0;

ENDDATA

! Minimize shipping cost over all links;

MIN = @SUM( NET(I, J, K): C(I, J) * X(I, J, K));

! This is the balance constraint. There are two cases:

Either the node that needs to be balanced is not a supply,

in which case the sum of incoming amounts

minus the sum of outgoing amounts must equal

the demand for that commodity for that city;

!or where the node is a supply,

the sum of incoming minus outgoing amounts must equal

the negative of the sum of the demand for the commodity

that the node supplies;

@FOR(COMMO(K): @FOR(NODES(J)|J #NE# K:

@SUM(NODES(I): V(I, J) * X(I, J, K) - V(J, I) * X(J, I, K))

= D(K, J);

);

@FOR(NODES(J)|J #EQ# K:

@SUM(NODES(I): V(I, J) * X(I, J, K) - V(J, I) * X(J, I, K))

= -@SUM( NODES(L): D(K, L))););

! This is a capacity constraint;

@FOR(EDGES(I, J)|I #NE# J:

@SUM(COMMO(K): X(I, J, K)) <= U(I, J);

);

END

Notice there are 3 (commodities) 6 (cities) = 18 balance constraints. If we instead identified

goods by origin/destination combination rather than just origin, there would be 9 6 = 54 balance

constraints. Solving, we get:




X( 1, 2, 1) 2.000000 0.0000000

X( 1, 3, 1) 3.000000 0.0000000

X( 1, 4, 1) 2.000000 0.0000000

X( 1, 5, 1) 1.000000 0.0000000

X( 1, 6, 1) 17.00000 0.0000000

X( 2, 3, 2) 2.000000 0.0000000

X( 2, 4, 2) 2.000000 0.0000000

X( 2, 5, 2) 1.000000 0.0000000

X( 3, 4, 5) 1.000000 0.0000000

X( 4, 2, 5) 4.000000 0.0000000

X( 4, 3, 2) 2.000000 0.0000000


X( 5, 3, 1) 1.000000 0.0000000

X( 5, 3, 5) 1.000000 0.0000000

X( 5, 4, 5) 5.000000 0.0000000

X( 5, 6, 5) 8.000000 0.0000000

X( 6, 2, 1) 3.000000 0.0000000

X( 6, 3, 1) 5.000000 0.0000000

X( 6, 4, 1) 5.000000 0.0000000

Notice, because of capacity limitations on other links, the depot city (6) is used for many of the

shipments.

8.9.4 Fleet Routing and Assignment An important problem in the airline and trucking industry is fleet routing and assignment. Given a set

of shipments or flights to be made, the routing part is concerned with the path each vehicle takes. This

is sometimes called the FTL(Full Truck Load) routing problem. The assignment part is of interest if

the firm has several different fleets of vehicles available. Then the question is what type of vehicle is

assigned to each flight or shipment. We will describe a simplified version of the approach used by

Subramanian et al. (1994) to do fleet assignment at Delta Airlines. A similar approach has been used at

US Airways by Kontogiorgis and Acharya (1999).

To motivate things, consider the following set of flights serving Chicago (ORD), Denver (DEN),

and Los Angeles (LAX) that United Airlines once offered on a typical weekday:

Daily Flight Schedule

City Time

Flight Depart Arrive Depart Arrive

1 221 ORD DEN 0800 0934

2 223 ORD DEN 0900 1039

3 274 LAX DEN 0800 1116

4 105 ORD LAX 1100 1314

5 228 DEN ORD 1100 1423

6 230 DEN ORD 1200 1521

7 259 ORD LAX 1400 1609

8 293 DEN LAX 1400 1510

9 412 LAX ORD 1400 1959

10 766 LAX DEN 1600 1912

11 238 DEN ORD 1800 2121

This schedule can be represented by the network in Figure 8.9. The diagonal lines from upper left

to lower right represent flight arrivals. The diagonal lines from lower left to upper right represent

departures. To complete the diagram, we need to add the lines connecting each flight departure to each

flight arrival. The thin line connecting the departure of Flight 274 from LAX to the arrival of Flight

274 in Denver illustrates one of the missing lines. If the schedule repeats every day, it is reasonable to

have the network have a backloop for each city, as illustrated for LAX. To avoid clutter, these lines

have not been added.


Figure 8.9 A Fleet Routing Network

Perhaps the obvious way of interpreting this as a network problem is as follows:

a) Each diagonal line (with the connection to its partner) constitutes a variable,

corresponding to a flight;

b) each horizontal line or backloop corresponds to a decision variable representing the

number of aircraft on the ground;

c) each point of either an arrival or a departure constitutes a node, and the model will have a

constraint saying, in words:

(no. of aircraft on the ground at this city at this instant) + (arrivals at this instant)

= (no. of departures from this city at this instant) + (no. of aircraft on the ground

after this instant).

With this convention, there would be 22 constraints (8 at ORD, 8 at DEN, and 6 at LAX), and 33

variables (11 flight variables and 22 ground variables). The number of constraints and variables can be

reduced substantially if we make the observation that the feasibility of a solution is not affected if, for

each city:

a) Each arrival is delayed until the first departure after that arrival.

b) Each departure is advanced (made earlier) to the most recent departure just after an

arrival.

Thus, the only nodes required are when a departure immediately follows an arrival.

If we have a fleet of just one type of aircraft, we probably want to know what is the minimum

number of aircrafts needed to fly this schedule. In words, our model is:

Minimize number of aircraft on the ground overnight

(That is the only place they can be, given the flight schedule)

subject to

source of aircraft = use of aircraft at each node of the network

and each flight must be covered.


Taking all the above observations into account gives the following formulation of a network LP.

Note the G variables represent the number of aircraft on the ground at a given city just after a specified

instant:

! Fleet routing with a single plane type;

! Minimize number of planes on ground overnight;

MIN = GC2400 + GD2400 + GL2400;

! The plane(old) conservation constraints;

! Chicago at 8 am, sources - uses = 0;

GC2400 - F221 - F223 - F105 - F259 - GC1400 = 0;

! Chicago at midnight;

GC1400 + F228 + F230 + F412 + F238 - GC2400 = 0;

! Denver at 11 am;

GD2400 + F221 + F223 - F228 - GD1100 = 0;

! Denver at high noon;

GD1100 + F274 - F230 - F293 - F238 - GD1800 = 0;

! Denver at midnight;

GD1800 + F766 - GD2400 = 0;

! LA at 8 am;

GL2400 - F274 - GL0800 = 0;

! LA at 1400;

GL0800 + F105 - F412 - GL1400 = 0;

! LA at 1600;

GL1400 + F293 - F766 - GL1600 = 0;

! LA at midnight;

GL1600 + F259 - GL2400 = 0;

! Cover our flight's constraints;

F221 = 1;

F223 = 1;

F274 = 1;

F105 = 1;

F228 = 1;

F230 = 1;

F259 = 1;

F293 = 1;

F412 = 1;

F766 = 1;

F238 = 1;


This model assumes no deadheading is used. That is, no plane is flown empty from one city to

another in order to position it for the next day. The reader probably figured out by simple intuitive

arguments that six aircraft are needed. The following solution gives the details:




GC2400 4.000000 0.0000000

GD2400 1.000000 0.0000000

GL2400 1.000000 0.0000000

F221 1.000000 0.0000000

F223 1.000000 0.0000000

F105 1.000000 0.0000000

F259 1.000000 0.0000000

GC1400 0.0000000 1.000000

F228 1.000000 0.0000000

F230 1.000000 0.0000000

F412 1.000000 0.0000000

F238 1.000000 0.0000000

GD1100 2.000000 0.0000000

F274 1.000000 0.0000000

F293 1.000000 0.0000000

GD1800 0.0000000 1.000000

F766 1.000000 0.0000000

GL0800 0.0000000 0.0000000

GL1400 0.0000000 0.0000000

GL1600 0.0000000 1.000000

Thus, there are four aircraft on the ground overnight at Chicago, one overnight at Denver, and one

overnight at Los Angeles.

8.9.5 Fleet Assignment If we have two or more aircraft types, then we have the additional decision of specifying the type of

aircraft assigned to each flight. The typical setting is we have a limited number of new aircraft that are

more efficient than previous aircraft. Let us extend our previous example by assuming we have two

aircraft of type B. They are more fuel-efficient than our original type A aircraft. However, the capacity

of type B is slightly less than A. We now probably want to maximize the profit contribution. The profit

contribution from assigning an aircraft of type i to flight j is:

+ (revenue from satisfying all demand on flight j)

(“spill” cost of not being able to serve all demand on j because of the limited capacity of

aircraft type i)

(the operating cost of flying aircraft type i on flight j)

+ (revenue from demand spilled from previous flights captured on this flight).

The spill costs and recoveries are probably the most difficult to estimate.

The previous model easily generalizes with the two modifications:

a) Conservation of flow constraints is needed for each aircraft type.

b) The flight coverage constraints become more flexible, because there are now two ways of

covering a flight.


After carefully calculating the profit contribution for each combination of aircraft type and flight,

we get the following model:

! Fleet routing and assignment with two plane types;

! Maximize profit contribution from flights covered;

MAX = 105 * F221A + 121 * F221B + 109 * F223A + 108

* F223B + 110 * F274A + 115 * F274B + 130 *

F105A + 140 * F105B + 106 * F228A + 122 *

F228B + 112 * F230A + 115 * F230B + 132 *

F259A + 129 * F259B + 115 * F293A + 123 *

F293B + 133 * F412A + 135 * F412B + 108 *

F766A + 117 * F766B + 116 * F238A + 124 *

F238B;

! Conservation of flow constraints;

! for type A aircraft;

! Chicago at 8 am, sources - uses = 0;

F221A - F223A - F105A - F259A - GC1400A + GC2400A=0;


F228A + F230A + F412A + F238A + GC1400A - GC2400A=0;

! Denver at 11 am;

F221A + F223A - F228A - GD1100A + GD2400A = 0;


F274A - F230A - F293A - F238A + GD1100A - GD1800A=0;


F766A - GD2400A + GD1800A = 0;

! LA at 8 am;

- F274A - GL0800A + GL2400A = 0;

! LA at 1400;

F105A - F412A + GL0800A - GL1400A = 0;

! LA at 1600;

F293A - F766A + GL1400A - GL1600A = 0;

! LA at midnight;

F259A - GL2400A + GL1600A = 0;

! Aircraft type B, conservation of flow;

! Chicago at 8 am;

-F221B - F223B - F105B - F259B - GC1400B +GC2400B=0;


F228B + F230B + F412B + F238B + GC1400B - GC2400B=0;

! Denver at 11 am;

F221B + F223B - F228B - GD1100B + GD2400B = 0;


F274B - F230B - F293B - F238B + GD1100B - GD1800B=0;


F766B - GD2400B + GD1800B = 0;

! LA at 8 am;

- F274B - GL0800B + GL2400B = 0;

! LA at 1400;

F105B - F412B + GL0800B - GL1400B = 0;

! LA at 1600;

F293B - F766B + GL1400B - GL1600B = 0;

! LA at midnight;

F259B - GL2400B + GL1600B = 0;

! Can put at most one plane on each flight;


F221A + F221B <= 1;

F223A + F223B <= 1;

F274A + F274B <= 1;

F105A + F105B <= 1;

F228A + F228B <= 1;

F230A + F230B <= 1;

F259A + F259B <= 1;

F293A + F293B <= 1;

F412A + F412B <= 1;

F766A + F766B <= 1;

F238A + F238B <= 1;

! Fleet size of type B;

GC2400B + GD2400B + GL2400B <= 2;

The not so obvious solution is:




F221B 1.000000 0.0000000

F223A 1.000000 0.0000000

F274A 1.000000 0.0000000

F105A 1.000000 0.0000000

F228B 1.000000 0.0000000

F230A 1.000000 0.0000000

F259A 1.000000 0.0000000

F293B 1.000000 0.0000000

F412A 1.000000 0.0000000

F766B 1.000000 0.0000000

F238A 1.000000 0.0000000

GC2400A 3.000000 0.0000000

GD1100A 1.000000 0.0000000

GL2400A 1.000000 0.0000000

GC2400B 1.000000 0.0000000

GD1100B 1.000000 0.0000000

GD2400B 1.000000 0.0000000

Six aircraft are still used. The newer type B aircraft cover flights 221, 228, 293, and 766. Since

there are two vehicle types, this model is a multicommodity network flow model rather than a pure

network flow model. Thus, we are not guaranteed to be able to find a naturally integer optimal solution

to the LP. Nevertheless, such was the case for the example above.


Generating an explicit model as above would be tedious. The following is a set-based version of

the above model. With the set based version, adding a flight or an aircraft type is a fairly simple

clerical operation:

MODEL:

SETS: ! Fleet routing and assignment (FLEETRAV);

CITY :; ! The cities involved;

ACRFT: ! Aircraft types;

FCOST, ! Fixed cost per day of this type;

FSIZE; ! Max fleet size of this type;

FLIGHT:;

FXCXC( FLIGHT, CITY, CITY) :

DEPAT, ! Flight departure time;

ARVAT; ! arrival time at dest.;

AXC( ACRFT, CITY):

OVNITE; ! Number staying overnite by type,city;

AXF( ACRFT, FXCXC):

X, ! Number aircraft used by type,flight;

PC; ! Profit contribution by type,flight;

ENDSETS

DATA:

CITY = ORD DEN LAX;

ACRFT, FCOST, FSIZE =

MD90 0 7

B737 0 2;

FLIGHT = F221 F223 F274 F105 F228 F230 F259 F293 F412 F766 F238;

FXCXC, DEPAT, ARVAT =

! Flight Origin Dest. Depart Arrive;

F221 ORD DEN 800 934

F223 ORD DEN 900 1039

F274 LAX DEN 800 1116

F105 ORD LAX 1100 1314

F228 DEN ORD 1100 1423

F230 DEN ORD 1200 1521

F259 ORD LAX 1400 1609

F293 DEN LAX 1400 1510

F412 LAX ORD 1400 1959

F766 LAX DEN 1600 1912

F238 DEN ORD 1800 2121;

PC = ! Profit contribution of each vehicle*flight combo;

105 109 110 130 106 112

132 115 133 108 116

121 108 115 140 122 115

129 123 135 117 124;

ENDDATA

!-------------------------------------------------------------------;

! Maximize profit contribution from flights minus

overhead cost of aircraft in fleet;

MAX = @SUM( AXF( I, N, J, K): PC( I, N, J, K) * X( I, N, J, K))

- @SUM( AXC( I, J): FCOST( I) * OVNITE( I, J));


! At any instant, departures in particular, the number of

cumulative arrivals must be >= number of cumulative departures;

! For each flight of each aircraft type;

@FOR( ACRFT( I):

@FOR( FXCXC( N, J, K):

! Aircraft on ground in morning +

number aircraft arrived thus far >=

number aircraft departed thus far;

OVNITE( I, J) +

@SUM( FXCXC( N1, J1, K1)| K1 #EQ# J #AND#

ARVAT( N1, J1, K1) #LT# DEPAT( N, J, K):

X( I, N1, J1, J)) >=

@SUM( FXCXC( N1, J1, K1)| J1 #EQ# J #AND#

DEPAT( N1, J1, K1) #LE# DEPAT( N, J, K):

X( I, N1, J, K1));

););

! This model does not allow deadheading, so at the end of the day,

arrivals must equal departures;

@FOR( ACRFT( I):

@FOR( CITY( J):

@SUM( AXF( I, N, J1, J): X( I, N, J1, J)) =

@SUM( AXF( I, N, J, K): X( I, N, J, K));

);

);

! Each flight must be covered;

@FOR( FXCXC( N, J, K):

@SUM( AXF( I, N, J, K): X( I, N, J, K)) = 1;

);

! Fleet size limits;

@FOR( ACRFT( I):

@SUM( AXC( I, J): OVNITE( I, J)) <= FSIZE( I);

);

! Fractional planes are not allowed;

@FOR( AXF: @GIN( X); );

END

Sometimes, especially in trucking, one has the option of using rented vehicles to cover only

selected trips. With regard to the model, the major modification is that rented vehicles do not have to

honor the conservation of flow constraints. Other details that are sometimes included relate to

maintenance. With aircraft, for example, a specific aircraft must be taken out of service for

maintenance after a specified number of landings, or after a specified number of flying hours, or after a

certain elapsed time, whichever occurs first. It is not too difficult to incorporate such details, although

the model becomes substantially larger.

8.9.6 Leontief Flow Models In a Leontief flow model, each activity produces one output. However, it may use zero or more inputs.

The following example illustrates.

Example: Islandia Input-Output Model

The country of Islandia has four major export industries: steel, automotive, electronics, and plastics.

The economic minister of Islandia would like to maximize exports-imports. The unit of exchange in

Islandia is the klutz. The prices in klutzes on the world market per unit of steel, automotive,


electronics, and plastics are, respectively: 500, 1500, 300, and 1200. Production of one unit of steel

requires 0.02 units of automotive production, 0.01 units of plastics, 250 klutzes of raw material

purchased on the world market, plus one-half man-year of labor. Production of one automotive unit

requires 0.8 units of steel, 0.15 units of electronics, 0.11 units of plastic, one man-year of labor, and

300 klutzes of imported material. Production of one unit of electronic equipment requires 0.01 units of

steel, 0.01 units of automotive, 0.05 units of plastic, half a man-year of labor, and 50 klutzes of

imported material. Automotive production is limited at 650,000 units. Production of one unit of plastic

requires 0.03 units of automotive production, 0.2 units of steel, 0.05 units of electronics, 2 man-years

of labor, plus 300 klutzes of imported materials. The upper limit on plastic is 60,000 units. The total

manpower available in Islandia is 830,000 men per year. No steel, automotive, electronics, or plastic

products may be imported.

How much should be produced and exported of the various products?

Formulation and Solution of the Islandia Problem

The formulation of an input-output model should follow the same two-step procedure for formulating

any LP model. Namely, (1) identify the decision variables and (2) identify the constraints. The key to

identifying the decision variables for this problem is to make the distinction between the amount of

commodity produced and the amount exported. Once this is done, the decision variables can be

represented as:

PROD(STEEL) = units of steel produced,

PROD(AUTO) = units of automotive produced,

PROD(PLASTIC) = units of plastic produced,

PROD(ELECT) = units of electronics produced,

EXP(STEEL) = units of steel exported,

EXP(AUTO) = units of automotive exported,

EXP(PLASTIC) = units of plastic exported,

EXP(ELECT) = units of electronics exported.

The commodities can be straightforwardly identified as steel, automotive, electronics, plastics,

manpower, automotive capacity, and plastics capacity. Thus, there will be seven constraints.


The sets formulation and solution are:

MODEL: ! Islandia Input/output model;

SETS:

COMMO:

PROD, EXP, REV, COST, MANLAB, CAP;

CXC(COMMO, COMMO): USERATE;

ENDSETS

DATA:

COMMO = STEEL, AUTO, PLASTIC, ELECT;

COST = 250 300 300 50;

REV = 500 1500 1200 300;

MANLAB = .5 1 2 .5;

! Amount used of the column commodity per unit

of the row commodity;

USERATE= -1 .02 .01 0

.8 -1 .11 .15

.2 .03 -1 .05

.01 .01 .05 -1;

MANPOWER = 830000;

CAP = 999999 650000 60000 999999;

ENDDATA

[PROFIT] MAX = @SUM( COMMO: REV * EXP - PROD * COST);

@FOR( COMMO( I):

[ NETUSE] ! Net use must equal = 0;

EXP(I) + @SUM(COMMO(J): USERATE(J,I)* PROD(J))

= 0;

[CAPLIM] PROD( I) <= CAP( I);

);

[MANLIM] @SUM(COMMO:PROD * MANLAB) < MANPOWER;

END

Notice this model has the Leontief flow feature. Namely, each decision variable has at most one

negative constraint coefficient.

The solution is:

Global optimal solution found.

Objective value: 0.4354312E+09


MANPOWER 830000.0 0.000000

PROD( STEEL) 393958.3 0.000000

PROD( AUTO) 475833.3 0.000000

PROD( PLASTIC) 60000.00 0.000000

PROD( ELECT) 74375.00 0.000000

EXP( STEEL) 547.9167 0.000000

EXP( AUTO) 465410.4 0.000000

EXP( PLASTIC) 0.000000 2096.875

EXP( ELECT) 0.000000 121.8750

REV( STEEL) 500.0000 0.000000

REV( AUTO) 1500.000 0.000000

REV( PLASTIC) 1200.000 0.000000


REV( ELECT) 300.0000 0.000000

COST( STEEL) 250.0000 0.000000

COST( AUTO) 300.0000 0.000000

COST( PLASTIC) 300.0000 0.000000

COST( ELECT) 50.00000 0.000000

MANLAB( STEEL) 0.5000000 0.000000

MANLAB( AUTO) 1.000000 0.000000

MANLAB( PLASTIC) 2.000000 0.000000

MANLAB( ELECT) 0.5000000 0.000000

CAP( STEEL) 999999.0 0.000000

CAP( AUTO) 650000.0 0.000000

CAP( PLASTIC) 60000.00 0.000000

CAP( ELECT) 999999.0 0.000000

USERATE( STEEL, STEEL) -1.000000 0.000000

USERATE( STEEL, AUTO) 0.2000000E-01 0.000000

USERATE( STEEL, PLASTIC) 0.1000000E-01 0.000000

USERATE( STEEL, ELECT) 0.000000 0.000000

USERATE( AUTO, STEEL) 0.8000000 0.000000

USERATE( AUTO, AUTO) -1.000000 0.000000

USERATE( AUTO, PLASTIC) 0.1100000 0.000000

USERATE( AUTO, ELECT) 0.1500000 0.000000

USERATE( PLASTIC, STEEL) 0.2000000 0.000000

USERATE( PLASTIC, AUTO) 0.3000000E-01 0.000000

USERATE( PLASTIC, PLASTIC) -1.000000 0.000000

USERATE( PLASTIC, ELECT) 0.5000000E-01 0.000000

USERATE( ELECT, STEEL) 0.1000000E-01 0.000000

USERATE( ELECT, AUTO) 0.1000000E-01 0.000000

USERATE( ELECT, PLASTIC) 0.5000000E-01 0.000000

USERATE( ELECT, ELECT) -1.000000 0.000000


PROFIT 0.4354312E+09 1.000000

NETUSE( STEEL) 0.000000 500.0000

CAPLIM( STEEL) 606040.7 0.000000

NETUSE( AUTO) 0.000000 1500.000

CAPLIM( AUTO) 174166.7 0.000000

NETUSE( PLASTIC) 0.000000 3296.875

CAPLIM( PLASTIC) 0.000000 2082.656

NETUSE( ELECT) 0.000000 421.8750

CAPLIM( ELECT) 925624.0 0.000000

MANLIM 0.000000 374.0625

The solution indicates the best way of selling Islandia’s steel, automotive, electronics, plastics, and

manpower resources is in the form of automobiles.

This problem would fit the classical input-output model format of Leontief if, instead of

maximizing profits, target levels were set for the export (or consumption) of steel, automotive, and

plastics. The problem would then be to determine the production levels necessary to support the

specified export/consumption levels. In this case, the objective function is irrelevant.

A natural generalization is to allow alternative technologies for producing various commodities.

These various technologies may correspond to the degree of mechanization or the form of energy

consumed (e.g., gas, coal, or hydroelectric).


8.9.7 Activity/Resource Diagrams The graphical approach for depicting a model can be extended to arbitrary LP models. The price one

must pay to represent a general LP graphically is one must introduce an additional component type into

the network. There are two component types in such a diagram: (1) activities, which correspond to

variables and are denoted by a square, and (2) resources, which correspond to constraints and are

denoted by a circle. Each constraint can be thought of as corresponding to some commodity and, in

words, as saying “uses of commodity sources of commodity”. The arrows incident to a square

correspond to the resources, commodities, or constraints with which that variable has an interaction.

The arrows incident to a circle must obviously then correspond to the activities or decision variables

with which the constraint has an interaction.

Example: The Vertically Integrated Farmer

A farmer has 120 acres that can be used for growing wheat or corn. The yield is 55 bushels of wheat or

95 bushels of corn per acre per year. Any fraction of the 120 acres can be devoted to growing wheat or

corn. Labor requirements are 4 hours per acre per year, plus 0.15 hours per bushel of wheat, and 0.70

hours per bushel of corn. Cost of seed, fertilizer, etc., is 20 cents per bushel of wheat produced and 12

cents per bushel of corn produced. Wheat can be sold for $1.75 per bushel and corn for $0.95 per

bushel. Wheat can be bought for $2.50 per bushel and corn for $1.50 per bushel.

In addition, the farmer may raise pigs and/or poultry. The farmer sells the pigs or poultry when

they reach the age of one year. A pig sells for $40. He measures the poultry in terms of coops. One

coop brings in $40 at the time of sale. One pig requires 25 bushels of wheat or 20 bushels of corn, plus

25 hours of labor and 25 square feet of floor space. One coop of poultry requires 25 bushels of corn or

10 bushels of wheat, plus 40 hours of labor and 15 square feet of floor space.

The farmer has 10,000 square feet of floor space. He has available per year 2,000 hours of his own

time and another 2,000 hours from his family. He can hire labor at $1.50 per hour. However, for each

hour of hired labor, 0.15 hour of the farmer’s time is required for supervision. How much land should

be devoted to corn and to wheat, and how many pigs and/or poultry should be raised to maximize the

farmer’s profits? This problem is based on an example in chapter 12 of Hadley (1962).


You may find it convenient to use the following variables for this problem:

WR Wheat raised (in bushels)

CR Corn harvested (in bushels)

PS Pigs raised and sold

HS Hens raised and sold (number of coops)

LB Labor hired (in hours)

WS Wheat marketed or sold (in bushels)

CS Corn marketed or sold (in bushels)

CH Corn used to feed hens (in bushels)

WH Wheat used to feed hens (in bushels)

CP Corn used to feed pigs (in bushels)

WP Wheat used to feed pigs (in bushels)

CB Corn bought (in bushels)

WB Wheat bought ( in bushels)

The activity-resource diagram for the preceding problem is shown in Figure 8.10:

Figure 8.10 An Activity-Resource Diagram

Some things to note about an activity-resource diagram are:

Each rectangle in the diagram corresponds to a decision variable in the formulation.

Each circle in the diagram corresponds to a constraint or the objective.

Each arrow in the diagram corresponds to a coefficient in the formulation.

Associated with each circle or rectangle is a unit of measure (e.g., hours or bushels).

The units or dimension of each arrow is:

“Units of the circle” per “unit of the rectangle.”

1.5

.12


Below is the formulation corresponding to the above diagram.

! All constraints are in Uses <= Sources form;

[PROFIT] MAX= 1.75*WS +.95*CS +40*PS +40*HS -1.5*LB -.2*WR -.12*CR -1.5*CB -2.5*WB;

[LAND] (1/55)*WR + (1/95)*CR <= 120 ; ! Acres;

[LABOR] (.15+4/55)*WR + (.7+4/95)*CR + 40*HS + 25*PS <= .85*LB + 4000 ; ! Hours;

[HRDLABOR] LB <= 2000/.15; ! Hours;

[WHEAT] WS + WH + WP <= WB + WR; ! Bushels);

[CORN] CS + CH + CP <= CB + CR; ! Bushels;

[HENFEED] HS <= (1/25)*CH + (1/10)*WH; ! Coops;

[PIGFEED] PS <= (1/20)*CP + (1/25)*WP; ! Pigs;

[FLOORSP] 25*PS + 15*HS <= 10000; ! Square feet;

Notice that there is a one-to-one correspondence between the rows of the formulation and the round

nodes of the diagram, and a one-to-one correspondence between the variables of the formulation and

the square “hyper-arcs” of the diagram. The solution is:


WS 5967.500 0.000000

CS 0.000000 0.4122697

PS 0.000000 0.000000

HS 63.25000 0.000000

LB 0.000000 1.021875

WR 6600.000 0.000000

CR 0.000000 0.000000

CB 0.000000 0.1377303

WB 0.000000 0.7500000

WH 632.5000 0.000000

WP 0.000000 0.7125000

CH 0.000000 0.6622697

CP 0.000000 0.0653947


PROFIT 11653.12 1.000000

LAND 0.000000 78.35938

LABOR 0.000000 0.5625000

HRDLABOR 13333.33 0.000000

WHEAT 0.000000 1.750000

CORN 0.000000 1.362270

HENFEED 0.000000 17.50000

PIGFEED 0.000000 25.93750

FLOORSP 9051.250 0.000000

Notice that the most profitable use of land is to raise wheat. The most profitable use of the farmer’s

own labor and floor space is to use it, plus some wheat, to raise hens.

8.9.8 Spanning Trees Another simple yet important network-related problem is the spanning tree problem. It arises, for

example, in the installation of utilities such as cable, power lines, roads, and sewers to provide services

to homes in newly developed regions. Given a set of homes to be connected, we want to find a

minimum cost network, so every home is connected to the network. A reasonable approximation to the

cost of the network is the sum of the costs of the arcs in the network. If the arcs have positive costs,

then a little reflection should convince you the minimum cost network contains no loops (i.e., for any


two nodes (or homes) on the network, there is exactly one path connecting them). Such a network is

called a spanning tree.

A simple algorithm is available for finding a minimal cost spanning tree, see Kruskal (1956):

1. Set Y = {2, 3, 4 ... n} (i.e., the set of nodes yet to be connected).

A = {1} (i.e., the set of already connected nodes). We may arbitrarily define node 1 as the

root of the tree.

2. If Y is empty, then we are done,

3. else find the shortest arc (i,j) such that i is in A and j is in Y.

4. Add arc (i, j) to the network and

set A = A + j,

Y = Y j.

5. Go to (2).

Because of the above simple and efficient algorithm, LP is not needed to solve the minimum

spanning tree problem. In fact, formulating the minimum spanning tree problem as an LP is a bit

tedious.

The following illustrates a LINGO model for a spanning tree. This model does not explicitly solve

it as above, but just solves it as a straightforward integer program:

MODEL: ! (MNSPTREE);

!Given a set of nodes and the distance between each pair, find

the shortest total distance of links on the network to connect

all the nodes. This is the classic minimal spanning tree (MST)

problem;

SETS:

CITY: LVL;

! LVL( I) = level of city I in tree. LVL( 1) = 0;

LINK( CITY, CITY):

DIST, ! The distance matrix;

X; ! X( I,J) = 1 if we use link I, J;

ENDSETS

! This model finds the minimum cost network connecting Atlanta,

Chicago, Cincinnati, Houston, LA, and Montreal so that

messages can be sent from Atlanta (base) to all other cities;

DATA:

CITY= ATL CHI CIN HOU LAX MON;

! Distance matrix need not be symmetric. City 1 is base;

DIST = 0 702 454 842 2396 1196 !from Atlanta;

702 0 324 1093 2136 764 !from Chicago;

454 324 0 1137 2180 798 !from Cinci;

842 1093 1137 0 1616 1857 !from Houston;

2396 2136 2180 1616 0 2900 !from LA;

1196 764 798 1857 2900 0;!from Montreal;

ENDDATA

!----------------------------------------------;

!The model size: Warning, may be slow for N > 8;

N = @SIZE( CITY);

!The objective is to minimize total dist. of links;

MIN = @SUM( LINK: DIST * X);


!For city K, except the base, ... ;

@FOR( CITY( K)| K #GT# 1: ! It must be entered;

@SUM( CITY( I)| I #NE# K: X( I, K)) = 1;

!If there is a link from J-K, then LVL(K)=LVL(J)+1.

Note:These are not very powerful for large problems;

@FOR( CITY( J)| J #NE# K:

LVL( K) >= LVL( J) + X( J, K)

- ( N - 2) * ( 1 - X( J, K))

+ ( N - 3) * X( K, J); ); );

LVL( 1) = 0; ! City 1 has level 0;

!There must be an arc out of city 1;

@SUM( CITY( J)| J #GT# 1: X( 1, J)) >= 1;

!Make the X's 0/1;

@FOR( LINK: @BIN( X); );

!The level of a city except the base is at least 1 but no more than

N-1, and is 1 if link to the base;

@FOR( CITY( K)| K #GT# 1:

@BND( 1, LVL( K), 999999);

LVL( K) <= N - 1 - ( N - 2) * X( 1, K); );

END

The solution is:




N 6.000000 0.0000000

LVL( CHI) 2.000000 0.0000000

LVL( CIN) 1.000000 0.0000000

LVL( HOU) 1.000000 0.0000000

LVL( LAX) 2.000000 0.0000000

LVL( MON) 3.000000 0.0000000

X( ATL, CIN) 1.000000 454.0000

X( ATL, HOU) 1.000000 842.0000

X( CHI, MON) 1.000000 764.0000

X( CIN, CHI) 1.000000 324.0000

X( HOU, LAX) 1.000000 1616.000

The solution indicates Atlanta should connect to Cincinnati and Houston. Houston, in turn

connects to LA. Cincinnati connects to Chicago, and Chicago connects to Montreal.

8.9.9 Steiner Trees A Steiner tree is a generalization of a minimal spanning tree. The difference is, for a given network,

only a specified subset of the nodes need be connected in a Steiner tree. Providing network services in

a new housing development is a simple example, such as communication cable, sewer lines, water

lines, and roads. Each house must be connected to the network, but not all possible nodes or junctions

in the candidate network need be included.

Example

The first computer, an IBM RS6000, to beat a grandmaster, Gary Kasparov, at the game of chess,

contained electronic chips designed with the help of Steiner-tree-like optimization methods. A typical

VLSI (Very Large Scale Integrated) chip on this computer was less than 2 millimeters on a side.


Nevertheless, it might contain over 120 meters of connecting pathways for connecting the various

devices on the chip. An important part of increasing the speed of a chip is reducing the length of the

paths on the chip. Figure 8.11 shows a chip on which five devices must be connected on a common

tree. Because of previous placement of various devices, the only available links are the ones indicated

in the figure. The square nodes, A, B, C, D, and E, must be connected. The round nodes, F, G, etc.,

may be used, but need not be connected. What set of links should be used to minimize the total

distance of links?

Figure 8.11 Steiner Tree Problem

A

B

C

D

E

F G H

I J

K

Finding a minimal length Steiner tree is considerably harder than finding a minimal length

spanning tree. For small problems, the following LINGO model will find minimal length Steiner trees.

The data correspond to the network in Figure 8.11:

MODEL: ! (STEINERT);

!Given a set of nodes, the distance between them, and a specified

subset of the nodes, find the set of links so that the total distance

is minimized, and there is a (unique) path between every pair of

nodes in the specified subset. This is called a Steiner tree problem;

SETS:

ALLNODE : U;

! U( I) = level of node I in the tree;

! U( 1) = 0;

MUSTNOD( ALLNODE); ! The subset of nodes that must be connected;

LINK( ALLNODE, ALLNODE):

DIST, ! The distance matrix;

X; ! X( I, J) = 1 if we use link I, J;

ENDSETS


DATA:

ALLNODE= ! Distance matrix need not be symmetric;

A B C D E F G H I J K;

DIST =0 14 999 999 999 4 999 999 999 999 999

14 0 999 999 999 999 999 3 999 999 999

999 999 0 9 999 999 999 999 2 999 999

999 999 9 0 999 999 999 999 999 3 6

999 999 999 999 0 999 999 5 999 999 3

4 999 999 999 999 0 999 999 3 999 999

999 999 999 999 999 999 0 2 999 3 999

999 3 999 999 5 999 2 0 999 999 999

999 999 2 999 999 3 999 999 0 8 999

999 999 999 3 999 999 3 999 8 0 999

999 999 999 6 3 999 999 999 999 999 0;

! The subset of nodes that must be connected.

The first node must be a must-do node;

MUSTNOD = A B C D E;

ENDDATA

!-----------------------------------------------;

! The model size: Warning, may be slow for N > 8;

N = @SIZE( ALLNODE);

! Objective is minimize total distance of links;

MIN = @SUM( LINK: DIST * X);

! For each must-do node K, except the base, ... ;

@FOR( MUSTNOD( K)| K #GT# 1:

! It must be entered;

@SUM( ALLNODE( I)| I #NE# K: X( I, K)) = 1;

! Force U(J)=number arcs between node J and node 1. Note: This is not

very strong for large problems;

@FOR( ALLNODE( J)| J #GT# 1 #AND# J #NE# K:

U( J) >= U( K) + X ( K, J) -

( N - 2) * ( 1 - X( K, J)) +

( N - 3) * X( J, K); );

);

! There must be an arc out of node 1;

@SUM( ALLNODE( J)| J #GT# 1: X( 1, J)) >= 1;

!If an arc out of node J, there must be an arc in;

@FOR( ALLNODE( J)| J #GT# 1:

@FOR( ALLNODE( K)| K #NE# J:

@SUM( ALLNODE( I)| I #NE# K #AND# I #NE# J:

X( I, J)) >= X( J, K);

); );

! Make the X's 0/1;

@FOR( LINK: @BIN( X); );

! Level of a node except the base is at least 1, no more than N-1,

and is 1 if link to the base;

@FOR( ALLNODE( K)| K #GT# 1:

@BND( 1, U( K), 999999);

U( K) < N - 1 - ( N - 2) * X( 1, K); );

END


The solution has a cost of 33. The list of links used is:



Branch count: 0


X( A, F) 1.000000 4.000000

X( F, I) 1.000000 3.000000

X( G, H) 1.000000 2.000000

X( H, B) 1.000000 3.000000

X( H, E) 1.000000 5.000000

X( I, C) 1.000000 2.000000

X( I, J) 1.000000 8.000000

X( J, D) 1.000000 3.000000

X( J, G) 1.000000 3.000000


The corresponding minimal length Steiner tree appears in Figure 8.12:

Figure 8.12 Minimal Length Steiner Tree

A

B

C

D

E

F G H

I J

Notice node K is not in the tree. This example is missing two important features of real chip

design: a) It shows only two-dimensional paths. In fact, three-dimensional paths are possible (and

typically needed) by adding vertical layers to the chip. b) It only shows one tree; whereas, in fact,

there may be many distinct trees to be constructed (e.g., some devices need to be connected to

electrical “ground”, others need to be connected to the clock signal, etc.). This might be handled by

solving the trees sequentially, the more complicated trees first.

8.10 Nonlinear Networks There are a number of important problems where the constraints describe a network problem.

However, either the objective is nonlinear, or there are additional conditions on the network that are

nonlinear. The first example describes a transportation problem where the value of shipping or

assigning an item to a destination depends upon (a) how many items have already been shipped to that

destination, and (b) the type of item and type of destination. In the military, this kind of problem is

known as a weapons or target assignment problem. If we define:

x(i,j) = number of units of type j assigned to task i;

p(i,j) = Prob{ a unit of type j will not successfully complete task i}.

Then, assuming independence, the probability task i will not be completed is:

p(i, 1)x(i, 1)

p(i, 2)x(i, 2)

… p(i, n)x(i, n)

.

The log of the proability that task i will not be completed is:

x(i, 1) *log[p(i, 1)]+ x(i, 2) *log[p(i, 2)]+ …+ x(i, n) *log[p(i, n)].

A reasonable objective is to maximize the expected value of successfully completed tasks. The

following model illustrates this idea, using data from Bracken and McCormick (1968):

MODEL:

! (TARGET) Bracken and McCormick;

SETS:

DESTN/1..20/: VALUE, DEM, LFAILP;

SOURCE/1..5/: AVAIL;


DXS( DESTN, SOURCE): PROB, VOL;

ENDSETS

DATA:

! Probability that a unit from source J will NOT do the job at

destination I;

PROB=

1.00 .84 .96 1.00 .92

.95 .83 .95 1.00 .94

1.00 .85 .96 1.00 .92

1.00 .84 .96 1.00 .95

1.00 .85 .96 1.00 .95

.85 .81 .90 1.00 .98

.90 .81 .92 1.00 .98

.85 .82 .91 1.00 1.00

.80 .80 .92 1.00 1.00

1.00 .86 .95 .96 .90

1.00 1.00 .99 .91 .95

1.00 .98 .98 .92 .96

1.00 1.00 .99 .91 .91

1.00 .88 .98 .92 .98

1.00 .87 .97 .98 .99

1.00 .88 .98 .93 .99

1.00 .85 .95 1.00 1.00

.95 .84 .92 1.00 1.00

1.00 .85 .93 1.00 1.00

1.00 .85 .92 1.00 1.00;

! Units available at each source;

AVAIL= 200 100 300 150 250;

! Min units required at each destination;

DEM=

30 0 0 0 0 100 0 0 0 40

0 0 0 50 70 35 0 0 0 10;

! Value of satisfying destination J;

VALUE=

60 50 50 75 40 60 35 30 25 150

30 45 125 200 200 130 100 100 100 150;

ENDDATA

!Max sum over I:(value of destn I)

*Prob{success at I};

MAX = @SUM( DESTN( I): VALUE( I) *

( 1 - @EXP( LFAILP( I))));

! The supply constraints;@FOR( SOURCE( J):

@SUM( DESTN( I): VOL( I, J)) <= AVAIL( J));

@FOR( DESTN( I):

!The demand constraints;

@SUM( SOURCE( J): VOL( I, J)) > DEM( I);

!Compute log of destination I failure probability;

@FREE( LFAILP( I));

LFAILP( I) =

@SUM(SOURCE(J): @LOG(PROB(I,J)) * VOL(I,J)););

END


Observe the model could be “simplified” slightly by using the equation computing LFAILP ( I) to

substitute out LFAILP ( I) in the objective. The time required to solve the model would probably

increase substantially, however, if this substitution were made. The reason is the number of variables

appearing nonlinearly in the objective increases dramatically. A general rule for nonlinear programs is:

If you can reduce the number of variables that appear nonlinearly in the objective or a

constraint by using linear constraints to define intermediate variables, then it is probably

worth doing.

Verify the constraint defining LFAILP (I) is linear in both LFAILP (I) and VOL ( I, J).

Notice the solution involves fractional assignments. A simple generalization of the model is to

require the VOL () variables to be general integers:




VOL( 1, 5) 50.81594 0.0000000

VOL( 2, 1) 13.51739 0.0000000

VOL( 2, 2) 1.360311 0.2176940

VOL( 2, 5) 45.40872 0.0000000

VOL( 3, 5) 48.62891 0.0000000

VOL( 4, 2) 23.48955 0.0000000

VOL( 5, 2) 20.89957 0.0000000

VOL( 6, 1) 100.0000 0.0000000

VOL( 7, 1) 39.10010 0.0000000

VOL( 8, 1) 27.06643 0.0000000

VOL( 8, 5) 0.4547474E-12 0.0000000

VOL( 9, 1) 20.31608 0.0000000

VOL( 10, 5) 51.13144 0.0000000

VOL( 11, 4) 33.19754 0.0000000

VOL( 12, 4) 40.93452 0.0000000

VOL( 13, 5) 54.01499 0.0000000

VOL( 14, 4) 58.82350 0.0000000

VOL( 14, 5) 0.5684342E-13 0.0000000

VOL( 15, 2) 26.21095 0.0000000

VOL( 15, 3) 43.78905 0.0000000

VOL( 16, 2) 24.23657 0.2176940

VOL( 16, 4) 17.04444 0.0000000

VOL( 17, 2) 3.803054 0.0000000

VOL( 17, 3) 72.03255 -0.4182476E-05

VOL( 18, 2) 0.8881784E-15 0.1489908

VOL( 18, 3) 57.55117 0.0000000

VOL( 19, 3) 64.21183 0.0000000

VOL( 20, 3) 62.41540 0.0000000


8.11 Problems 1. The Slick Oil Company is preparing to make next month’s pipeline shipment decisions. The Los

Angeles terminal will require 200,000 barrels of oil. This oil can be supplied from either Houston

or Casper, Wyoming. Houston can supply oil to L.A. at a transportation cost of $.25 per barrel.

Casper can supply L.A. at a transportation cost of $.28 per barrel. The St. Louis terminal will

require 120,000 barrels. St. Louis can be supplied from Houston at a cost of $.18 per barrel and

from Casper at a cost of $.22 per barrel. The terminal at Freshair, Indiana requires 230,000 barrels.

Oil can be shipped to Freshair from Casper at a cost of $.21 per barrel, from Houston at a cost of

$.19 per barrel, and from Titusville, Pa. at a cost of $.17 per barrel. Casper will have a total of

250,000 barrels available to be shipped. Houston will have 350,000 barrels available to be

shipped. Because of limited pipeline capacity, no more than 180,000 barrels can be shipped from

Casper to L.A. next month and no more than 150,000 barrels from Houston to L.A. The Newark,

N.J. terminal will require 190,000 barrels next month. It can be supplied only from Titusville at a

cost of $.14 per barrel. The Atlanta terminal will require 150,000 barrels next month. Atlanta can

be supplied from Titusville at a cost of $.16 per barrel or from Houston at a cost of $.20 per barrel.

Titusville will have a total of 300,000 barrels available to be shipped.

Formulate the problem of finding the minimum transportation cost distribution plan as a linear

program.

2. Louis Szathjoseph, proprietor of the Boulangerie Restaurant, knows he will need 40, 70, and 60

tablecloths on Thursday, Friday, and Saturday, respectively, for scheduled banquets. He can rent

tablecloths for three days for $2 each. A tablecloth must be laundered before it can be reused. He

can have them cleaned overnight for $1.50 each. He can have them laundered by regular one-day

service (e.g., one used on Thursday could be reused on Saturday) for $.80 each. There are

currently 20 clean tablecloths on hand with none dirty or at the laundry. Rented tablecloths need

not be cleaned before returning.

a) What are the decision variables?

b) Formulate the LP appropriate for minimizing the total cost of renting and laundering the

tablecloths. For each day, you will probably have a constraint requiring the number of

clean tablecloths available to at least equal that day’s demand. For each of the first two

days, you will probably want a constraint requiring the number of tablecloths sent to the

laundry not to exceed those that have been made dirty. Is it a network LP?

3. The Millersburg Supply Company uses a large fleet of vehicles it leases from manufacturers. The

following pattern of vehicle requirements is forecast for the next 8 months:

Month Jan Feb Mar Apr May Jun Jul Aug

Vehicles Required

430 410 440 390 425 450 465 470

Vehicles can be leased from various manufacturers at various costs and for various lengths of

time. The best plans available are: three-month lease, $1,700; four-month lease, $2,000;

five-month lease, $2,600. A lease can be started in any month. On January 1, there are 200 cars on

lease, all of which go off lease at the end of February.

a) Formulate an approach for minimizing Millersburg’s leasing costs over the 8 months.

b) Show that this problem is a network problem.


4. Several years ago, a university in the Westwood section of Los Angeles introduced a bidding

system for assigning professors to teach courses in its business school. The table below describes a

small, slightly simplified three-professor/two-course version. For the upcoming year, each

professor submits a bid for each course and places limits on how many courses he or she wants to

teach in each of the school’s two semesters. Each professor, however, is expected to teach four

courses total per year (at most three per semester).

Prof. X Prof. Y Prof. Z

Fall Courses 1 3 1

Spring Courses 3 2 3 Sections Needed in the Year

Min Max

Course A bids 6 3 8 3 7

Course B bids 4 7 2 2 8

From the table, note that: professor Z strongly prefers to teach course A; whereas, professor X

has a slight preference for A. Professor Y does not want to teach more than two course sections in

the Spring. Over both semesters, at least three sections of Course A must be taught. Can you

formulate this problem as a network problem?

5. Aircraft Fuel Ferrying Problem. Fuel cost is one of the major components of variable operating

cost for an airline. Some cities collect a tax on aircraft fuel sold at their airports. Thus, the cost per

liter of fuel may vary noticeably from one airport to another. A standard problem with any airliner

is the determination of how much fuel to take on at each stop. Fuel consumption is minimized if

just sufficient fuel is taken on at each stop to fly the plane to the next stop. This policy, however,

disregards the fact that fuel prices may differ from one airport to the next. Buying all the fuel at

the cheapest stop may not be the cheapest policy either. This might require carrying large fuel

loads that would in turn cause large amounts of fuel to be burned in ferrying the fuel. The

refueling considerations at a given stop on a route are summarized by the following three

numbers: (a) the minimum amount of fuel that must be on board at takeoff to make it to the next

stop, (b) the cost per liter of fuel purchased at this stop, and (c) the amount of additional fuel

above the minimum that is burned per liter of fuel delivered to the next stop. These figures are

given below for an airplane that starts at Dallas, goes to Atlanta, then Chicago, Des Moines, St.

Louis, and back to Dallas.

Dallas Atlanta Chicago Des Moines St. Louis

a) 3100 2700 1330 1350 2500

b) .29 .34 .35 .31 .33

c) .04 .03 .02 .01 .04

For example to fly from Dallas to Atlanta, the plane must take off with at least 3100 liters of

fuel. Any fuel purchased in Dallas costs $0.29 per liter. For each liter of fuel delivered to Atlanta

(i.e., still in the tank), an additional .04 liters had to be put in at Dallas. Alternatively, each

additional 1.04 liters loaded at Dallas, results in an additional liter delivered to Atlanta. The plane

has a maximum fuel carrying capacity of 6000 liters, which we will assume is independent of

airport. Also, assume the minimum amount of reserve fuel that must be on board for safety

reasons is fixed independent of airport, so we can act as if no reserve is required.


Formulate and solve a model for deciding how much fuel to buy at each airport. Is this

problem any form of a network LP?

6. Show that any LP can be converted to an equivalent LP in which every column (variable) has at

most three nonzero constraint coefficients. What does this suggest about the fundamental

complexity of a network LP vs. a general LP?

7. Figure 8.12 is the activity-on-arc diagram showing the precedence relations among the five

activities involved in repairing a refinery. The three numbers above each arc represent (from left to

right, respectively) the normal time for performing the activity in days, the time to perform the

activity if crashed to the maximum extent, and the additional cost in $1000s for each day the

activity is shortened. An activity can be partially crashed. It is desired the project be completed in

15 days. Write an LP formulation for determining how much each activity should be crashed.

Figure 8.12 PERT Diagram with Crashing Allowed

9, 5, 4

6, 3, 5

7, 2, 6

9, 4, 7

9, 2, 8

A

B

C

D

E

8. Given n currencies, the one-period currency exchange problem is characterized by a beginning

inventory vector, an exchange rate matrix, and an ending inventory requirement vector defined as

follows:

ni = amount of cash on hand in currency i, at the beginning of the period measured in units of

currency i, for i = 1, 2, ..., n;

rij = units of currency j obtainable per unit of currency i for i = 1, 2, ..., n, j = 1, 2, ..., n. Note

that rii = 1 and, in general, we can expect rij < 1/rji, for i j.

ei = minimum ending inventory requirement for currency i, for i = 1, 2, ..., n. That is, at the

end of the period, we must have at least ei units of currency i on hand.

The decision variables are:

Xij = amount of currency i converted into currency j, for i = 1, 2, ..., n; j = 1, 2, ..., n.


Formulate a model for determining an efficient set of values for the Xij. The formulation

should have the following features:

a) If there is a “money pump” kind of arbitrage opportunity, the model will find it.

b) It should not be biased against any particular currency (i.e., the solution should be

independent of which currency is called 1).

c) If a currency is worthless, you should buy no more of it than sufficient to meet the

minimum requirement. A currency i is worthless if rij = 0, for all j i.

9. The following linear program happens to be a network LP. Draw the corresponding network.

Label the nodes and links.

MIN = 4 * T + 2 * U + 3 * V + 5 * W + 6 * X + 7 * Y + 9 * Z;

[A] T + Y + Z >= 4;

[B] U - W - X - Z = 0;

[C] - T + W = 1;

[D] V + X - Y = 2;

[E] U + V <= 7;

END

10. Consider a set of three flights provided by an airline to serve four cities, A, B, C, and H. The

airline uses a two-fare pricing structure. The decision of how many seats or capacity to allocate to

each price class is sometimes called yield or revenue management. We would like to decide upon

how many seats to allocate to each fare class on each flight. Node H is a hub for changing planes.

The three flights are: from A to H, H to B, and H to C. The respective flight capacities are 120, 100,

and 110. Customer demand has the following characteristics:

Itinerary

Class 1 Demand

At a Price of

Class 2 Demand

At a Price of

AH 33 190 56 90

AB (via H) 24 244 43 193

AC (via H) 12 261 67 199

HB 44 140 69 80

HC 16 186 17 103

How many seats should be allocated to each class on each of the three flights? An obvious

solution, if it is feasible, is to set aside enough class 1 seats on every flight, so all class 1 travelers

can be accommodated. Thus, the leg AH would get 33 + 24 + 12 = 69 class 1 seats, leg HB would

get 24 + 44 = 68 class 1 seats and leg HC would get 12 + 16 = 28 class 1 seats. The total revenue

of this solution is $38,854. Is this the most profitable solution?


11. A common distribution system structure in many parts of the world is the three-level system

composed of plants, distribution centers (DC), and outlets. A cost minimization model for a

system composed of two plants (A & B), three DC’s (X, Y, and Z), and four outlets (1, 2, 3, and 4)

is shown below:

MIN = AX + 2 * AY + 3 * BX + BY + 2 * BZ + 5 * X1 +

7 * X2 + 9 * Y1 + 6 * Y2 + 7 * Y3 + 8 * Z2 + 7

* Z3 + 4 * Z4;

AX + AY = 9;

BX + BY + BZ = 8;

- AX - BX + X1 + X2 = 0;

- AY - BY + Y1 + Y2 + Y3 = 0;

- BZ + Z2 + Z3 + Z4 = 0;

- X1 - Y1 = - 3;

- X2 - Y2 - Z2 = - 5;

- Y3 - Z3 = - 4;

- Z4 = - 5;

END

Part of the solution is shown below:



AX 3.000000 0.0000000

AY 6.000000 0.0000000

BX 0.0000000 3.000000

BY 3.000000 0.0000000

BZ 5.000000 0.0000000

X1 3.000000 0.0000000

X2 0.0000000 0.0000000

Y1 0.0000000 5.000000

Y2 5.000000 0.0000000

Y3 4.000000 0.0000000

Z2 0.0000000 3.000000

Z3 0.0000000 1.000000

Z4 5.000000 0.0000000

a) Is there an alternate optimal solution to this distribution problem?

b) A trucking firm that offers services from city Y to city 1 would like to get more of your

business. At what price per unit might you be willing to give them more business according to

the above solution?

c) The demand at city 2 has been decreased to 3 units. Show how the model is changed.

d) The capacity of plant B has been increased to 13 units. Show how the model is changed.

e) Distribution center Y is actually in a large city where there is an untapped demand of 3 units

that could be served directly from the DC at Y. Show how to include this additional demand at

Y.


12. Labor on the first shift of a day (8 a.m. to 4 p.m.) costs $15 per person hour. Labor on the

second (4 p.m. to midnight) and third (midnight to 8 a.m.) shifts cost $20 per person hour and

$25 per person hour, respectively. A certain task requires 18 days if done with just first shift

labor and costs $8640. Second and third shift labor has the same efficiency as first shift labor. The

only way of accelerating or crashing the task is to add additional shifts for one or more additional

days. The total cost of the task consists solely of labor costs.

Complete the following crash cost table for this task.

Task time in whole days

18 17 16 15 14 13 12 11 10 9 8 7 6 5 4

Total cost

8640

13. You are on a camping trip and wish to prepare a recipe for a certain food delight that calls for 4

cups of water. The only containers in your possession are two ungraduated steel vessels, one of

3-cup capacity, the other of 5-cup capacity. Show how you can solve this problem by drawing a

certain two-dimensional network, where each node represents a specific combination of contents

in your two containers.

14. Following is part of the schedule for an airline:

City Time Difference

Flight Depart Arrive Depart Arrive in Profit

1 221 ORD DEN 0800 0934 +$3000

2 223 ORD DEN 0900 1039 $4000

3 274 LAX DEN 0800 1116 $3000

4 105 ORD LAX 1100 1314 +$10000

5 228 DEN ORD 1100 1423 $2000

6 230 DEN ORD 1200 1521 $3000

7 259 ORD LAX 1400 1609 +$4000

8 293 DEN LAX 1400 1510 +$1000

9 412 LAX ORD 1400 1959 +$7000

10 766 LAX DEN 1600 1912 +$2000

11 238 DEN ORD 1800 2121 $4000

The airline currently flies the above schedule using standard Boeing 737 aircraft. Boeing is

trying to convince the airline to use a new aircraft, the 737-XX, known affectionately as the Dos

Equis. The 737-XX consumes more fuel per kilometer. However, it is sufficiently larger such that,

if it carries enough passengers, it is more efficient per passenger kilometer. The “Difference in

Profit” column above shows the relative profitability of using the 737-XX instead of the standard

737 on each flight. The airline is considering using at most one 737-XX.

Based on the available information, analyze the wisdom of using the 737-XX in place of one

of the standard 737’s.


15. The following linear program happens to be a network LP:

MIN = 9 * S + 4 * T + 2 * U + 3 * V + 5 * W + 6 * X + 7 * Y;

[A] - T + W = 1;

[B] S + T + Y >= 4;

[C] U - W - X - S = 0;

[D] U + V <= 7;

[E] V + X - Y = 2;

END

a) Draw the corresponding network.

b) Label the nodes and links.

16. A small, but growing, long-distance phone company, SBG Inc., is trying to decide in which

markets it should try to expand. It has used the following model to decide how to maximize the

calls it carries per hour:

MAX = P1SEADNV + P2SEADNV + P1SEACHI + P2SEACHI

+ P1SEAATL + P2SEAATL + P1SEAMIA + P2SEAMIA

+ P1DNVCHI + P2DNVCHI + P1DNVATL + P2DNVATL

+ P1DNVMIA + P2DNVMIA + P1CHIATL + P2CHIATL

+ P1CHIMIA + P2CHIMIA + P1ATLMIA;

[LSEADNV] P1SEADNV + P2SEACHI + P2SEAATL + P2SEAMIA

+ P1DNVCHI + P2DNVATL + P2DNVMIA + P2CHIATL +

P2CHIMIA <= 95;

[LSEACHI] P2SEADNV + P1SEACHI + P1SEAATL + P1SEAMIA


P2CHIMIA <= 80;

[LDNVATL] P2SEADNV + P2SEACHI + P2SEAATL + P2SEAMIA


P2CHIMIA <= 200;

[LCHIATL] P2SEADNV + P2SEACHI + P1SEAATL + P1SEAMIA

+ P2DNVCHI + P2DNVALT + P2DNVMIA + P1CHIATL +

P1CHIMIA <= 110;

[LATLMIA] P1SEAMIA + P2SEAMIA + P1DNVMIA + P2DNVMIA

+ P1CHIMIA + P2CHIMIA + P1ATLMIA <= 105;

[DSEADNV] P1SEADNV + P2SEADNV <= 10;

[DSEACHI] P1SEACHI + P2SEACHI <= 20;

[DSEAATL] P1SEAATL + P2SEAATL <= 38;

[DSEAMIA] P1SEAMIA + P2SEAMIA <= 33;

[DDNVCHI] P1DNVCHI + P2DNVCHI <= 42;

[DDNVATL] P1DNVATL + P2DNVATL <= 48;

[DDNVMIA] P1DNVMIA + P2DNVMIA <= 23;

[DCHIATL] P1CHIATL + P2CHIATL <= 90;

[DCHIMIA] P1CHIMIA + P2CHIMIA <= 36;

[DATLMIA] P1ATLMIA <= 26;


Now, it would like to refine the model, so it takes into account not only revenue per call, but

also modest variable costs associated with each link for carrying a call. The variable cost per

typical call according to link used is shown in the table below:

Variable Cost/Call

DNV CHI ATL MIA

SEA .11 .16 X X

DNV X .15 X

CHI .06 X

ATL .07

An X means there is no direct link between the two cities. SBG would like to find the

combination of calls to accept to maximize profit contribution. Suppose the typical revenue per

call between ATL and SEA is $1.20. Show how to modify the model just to represent the revenue

and cost information for the demand between SEA and ATL.

17. Below is a four-activity project network presented in activity-on-node form, along with

information on crashing opportunities for each activity:

Activity Normal Time

(days)

Crash Cost Per

Day

Minimum Possible Time

(days)

A 8 3 4

B 7 4 5

C 6 6 3

D 9 2 5

Complete the following tabulation of crashing cost vs. project length:

Step Project Length

Incremental Crashing Cost/Day

Total Crashing

Cost

Activities to Crash

0 16 0 0 —

1

2

3

4

8 networks, distribution and pert/cpm€¦ · · 2016-12-248 networks, distribution and pert/cpm...

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